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1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
Bài 1:
a) 5(x-3)-4=2(x-1)
\(\Leftrightarrow5x-15-4=2x-2\)
\(\Leftrightarrow5x-19-2x+2=0\)
\(\Leftrightarrow3x-17=0\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)
Vậy: \(x=\frac{17}{3}\)
b) 5-(6-x)=4(3-2x)
\(\Leftrightarrow5-6+x=12-8x\)
\(\Leftrightarrow-1+x-12+8x=0\)
\(\Leftrightarrow-13+9x=0\)
\(\Leftrightarrow9x=13\)
\(\Leftrightarrow x=\frac{13}{9}\)
Vậy: \(x=\frac{13}{9}\)
c) (3x+5)(2x+1)=(6x-2)(x-3)
\(\Leftrightarrow6x^2+3x+10x+5=6x^2-18x-2x+6\)
\(\Leftrightarrow6x^2+13x+5=6x^2-20x+6\)
\(\Leftrightarrow6x^2+13x+5-6x^2+20x-6=0\)
\(\Leftrightarrow33x-1=0\)
\(\Leftrightarrow33x=1\)
\(\Leftrightarrow x=\frac{1}{33}\)
Vậy: \(x=\frac{1}{33}\)
d) \(\left(x+2\right)^2+2\left(x-4\right)=\left(x-4\right)\left(x-2\right)\)
\(\Leftrightarrow x^2+4x+4+2x-8=x^2-2x-4x+8\)
\(\Leftrightarrow x^2+6x-4=x^2-6x+8\)
\(\Leftrightarrow x^2+6x-4-x^2+6x-8=0\)
\(\Leftrightarrow12x-12=0\)
\(\Leftrightarrow x=1\)
Vậy:x=1
Bài 2:
a)\(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{x}{3}-\frac{5x}{6}-\frac{5x}{4}-\frac{x}{4}+5=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{10x}{12}-\frac{15x}{12}-\frac{3x}{12}+\frac{60}{12}=0\)
\(\Leftrightarrow4x-10x-15x-3x+60=0\)
\(\Leftrightarrow-24x+60=0\)
\(\Leftrightarrow-24x=-60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy: \(x=\frac{5}{2}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{3x-2}{2}-\frac{2x-1}{2}-\frac{x+3}{4}=0\)
\(\Leftrightarrow\frac{8x-3}{4}-\frac{2\left(3x-2\right)}{4}-\frac{2\left(2x-1\right)}{4}-\frac{x+3}{4}=0\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)-2\left(2x-1\right)-\left(x+3\right)=0\)
\(\Leftrightarrow8x-3-6x+4-4x+2-x-3=0\)
\(\Leftrightarrow-3x=0\)
\(\Leftrightarrow x=0\)
Vậy: x=0
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)}{30}-\frac{2\left(x+1\right)}{30}-\frac{5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow3x=-48\)
\(\Leftrightarrow x=-16\)
Vậy: x=-16
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}-\frac{1-x}{2}+2=0\)
\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}-\frac{12\left(1-x\right)}{24}+\frac{48}{24}=0\)
\(\Leftrightarrow9\left(3-x\right)+16\left(5-x\right)-12\left(1-x\right)+48=0\)
\(\Leftrightarrow27-9x+80-16x-12+12x+48=0\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow-13x=-143\)
\(\Leftrightarrow x=11\)
Vậy: x=11
e) \(\frac{3\left(5x-2\right)}{4}-2=\frac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\frac{3\left(5x-2\right)}{4}-2-\frac{7x}{3}+5\left(x-7\right)=0\)
\(\Leftrightarrow\frac{9\left(5x-2\right)}{12}-\frac{24}{12}-\frac{28x}{12}+\frac{60\left(x-7\right)}{12}=0\)
\(\Leftrightarrow9\left(5x-2\right)-24-28x+60\left(x-7\right)=0\)
\(\Leftrightarrow45x-18-24-28x+60x-420=0\)
\(\Leftrightarrow77x-462=0\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
Vậy:x=6
Bài 3:
a) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left(5x-4\right)\cdot2\cdot\left(2x+3\right)=0\)
Vì \(2\ne0\)
nên \(\left[{}\begin{matrix}5x-4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=\frac{-3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{4}{5};-\frac{3}{2}\right\}\)
b) \(\left(x-5\right)\left(3-2x\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\\x=\frac{-4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{5;\frac{3}{2};\frac{-4}{3}\right\}\)
c) \(\left(2x+1\right)\left(x^2+2\right)=0\)
Ta có: \(\left(2x+1\right)\left(x^2+2\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+2\ge2\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(2x+1=0\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy: \(x=\frac{-1}{2}\)
d) \(\left(8x-4\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow4\left(2x-1\right)\left(x^2+2x+2\right)=0\)
Ta có: \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta lại có \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1\ne0\forall x\)(3)
Ta có: \(4\ne0\)(4)
Từ (3) và (4) suy ra
2x-1=0
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
Bài 4:
a) \(\left(x-2\right)\left(2x+3\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2x^2+3x-4x-6=x^2-2x-x+2\)
\(\Leftrightarrow2x^2-x-6=x^2-3x+2\)
\(\Leftrightarrow2x^2-x-6-x^2+3x-2=0\)
\(\Leftrightarrow x^2+2x-8=0\)
\(\Leftrightarrow x^2+2x+1-9=0\)
\(\Leftrightarrow\left(x+1\right)^2-3^2=0\)
\(\Leftrightarrow\left(x+1-3\right)\left(x+1+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-4\right\}\)
b) \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)-\left(x-5\right)\left(4-x\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(2x+5+x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\cdot3x=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4\right\}\)
c) \(9x^2-1=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left[\left(3x-1\right)-\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(3x+1\right)\left(3x-1-2x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};-2\right\}\)
d) \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow x^2+4x+4-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2+4x+4-9x^2+36x-36=0\)
\(\Leftrightarrow-8x^2+40x-32=0\)
\(\Leftrightarrow-\left(8x^2-40x+32\right)=0\)
\(\Leftrightarrow-8\left(x^2-5x+4\right)=0\)
Vì \(-8\ne0\)
nên \(x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy: \(x\in\left\{1;4\right\}\)
e) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow4\left(4x^2+28x+49\right)-9\left(x^2+6x+9\right)=0\)
\(\Leftrightarrow16x^2+112x+196-9x^2-54x-81=0\)
\(\Leftrightarrow7x^2+58x+115=0\)
\(\Leftrightarrow7x^2+23x+35x+115=0\)
\(\Leftrightarrow x\left(7x+23\right)+5\left(7x+23\right)=0\)
\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-23\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-23}{7};-5\right\}\)
Bài 5:
a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left[\left(3x-2\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-2\\x=-1\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{2}{3};-1;\frac{1}{2}\right\}\)
b) \(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow2x^2-2x+x^2+2x-3=0\)
\(\Leftrightarrow3x^2-3=0\)
\(\Leftrightarrow3\left(x^2-1\right)=0\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)=0\)
Vì \(3\ne0\)
nên \(\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{1;-1\right\}\)
c) \(x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x^2-x+1\right)=0\)(5)
Ta có: \(x^2-x+1=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta lại có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\ne0\forall x\)(6)
Từ (5) và (6) suy ra
\(\left(x+1\right)^2=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy: x=-1
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
1)⇔x2+1x-3x+3=0
⇔x(x+1)-3(x+1)=0
⇔(x+1)(x-3)=0
⇔x+1=0 hoặc x-3=0
⇔x=-1 hoặc x=3
4)⇔x(1+5x)=0
⇔x=0 hoặc 1+5x=0
⇔x=0 hoặc 5x=-1
⇔x=0 hoặc x=-0.2
\(a,\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(3x-2-x+1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{2}{3};-1;\dfrac{1}{2}\right\}\)
\(b,\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x^2\right)=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)\left(1+x\right)-\left(1-x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1-x-x-3\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(-3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\-3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-1\right\}\)
\(c,\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x+2\right)\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\-5x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=\dfrac{7}{5}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;-2;\dfrac{7}{5}\right\}\)
\(d,x^4+x^3+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
Vậy phương trình có nghiệm duy nhất x = -1
\(e,x^3-7x+6=0\)
\(\Leftrightarrow x^3-4x-3x+6=0\)
\(\Leftrightarrow x\left(x^2-4\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3x-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{1;2;-3\right\}\)
\(f,x^4-4x^3+12x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)-\left(4x^3-12x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)-4x\left(x^2+3\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x^2-3-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3>0\forall x\\x^2-4x-3>0\forall x\end{matrix}\right.\)
Vậy phương trình vô nghiệm
\(g,x^5-5x^3+4x=0\)
\(\Leftrightarrow x\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^4-4x^2-x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^2-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\) hoặc x = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\\x=-1\end{matrix}\right.\) hoặc x =0
Vậy tập nghiệm của pt \(S=\left\{0;1;-1;2;-2\right\}\)
\(h,x^4-4x^3+3x^2+4x-4=0\)
\(\Leftrightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow\left(x^4-x^2\right)-\left(4x^3-4x\right)+\left(4x^2-4\right)=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4x\left(x^2-1\right)+4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{1;-1;2\right\}\)
3) \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
S=\(\left\{6;1\right\}\)
\(\)