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Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
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 = \(\left\{3;\dfrac{-1}{2}\right\}\)
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 \(-2\)
Ví dụ cho bạn một bài, còn lại tương tự.
a)Ta có: \(3x^4-5x^3+8x^2-5x+3\)
\(=3x^2\left(x-\frac{5}{6}\right)^2+\frac{71}{12}\left(x-\frac{30}{71}\right)^2+\frac{138}{71}>0\)
Vậy phương trình vô nghiệm.
a, \(x^4-5x^3+2x^2+10x+2=0\)
\(\Rightarrow x^4+x^3-6x^3-6x^2+8x^2+8x+2x+2=0\)
\(\Rightarrow x^3\left(x+1\right)-6x^2\left(x+1\right)+8x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)
Vì \(x^3-6x^2+8x+2>0\) nên \(x+1=0\Rightarrow x=-1\)
Các câu còn lại tương tự!
Chúc bạn học tốt!!!
a, \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Rightarrow5x^2-4x^2+8x-4-5=0\)
\(\Rightarrow x^2-x+9x-9=0\)
\(\Rightarrow x\left(x-1\right)+9\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\)
b, \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
\(\Rightarrow\left(x^2-9-x+3\right)\left(x^2-9+x-3\right)=0\)
\(\Rightarrow\left(x^2-x-6\right)\left(x^2+x-12\right)=0\)
\(\Rightarrow\left(x^2-3x+2x-6\right)\left(x^2+4x-3x-12\right)=0\)
\(\Rightarrow\left[x\left(x-3\right)+2\left(x-3\right)\right]\left[x\left(x+4\right)-3\left(x+4\right)\right]=0\)
\(\Rightarrow\left(x-3\right)\left(x+2\right)\left(x+4\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)
c, \(x^3-3x+2=0\)
\(\Rightarrow x^3+2x^2-2x^2-4x+x+2=0\)
\(\Rightarrow x^2\left(x+2\right)-2x\left(x+2\right)+\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x^2-2x+1\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
a, 5x2−4(x2−2x+1)−5=05x2−4(x2−2x+1)−5=0
⇒5x2−4x2+8x−4−5=0⇒5x2−4x2+8x−4−5=0
⇒x2−x+9x−9=0⇒x2−x+9x−9=0
⇒x(x−1)+9(x−1)=0⇒x(x−1)+9(x−1)=0
⇒(x−1)(x+9)=0⇒(x−1)(x+9)=0
⇒[x−1=0x+9=0⇒[x=1x=−9⇒[x−1=0x+9=0⇒[x=1x=−9
b, (x2−9)2−(x−3)2=0(x2−9)2−(x−3)2=0
⇒(x2−9−x+3)(x2−9+x−3)=0⇒(x2−9−x+3)(x2−9+x−3)=0
⇒(x2−x−6)(x2+x−12)=0⇒(x2−x−6)(x2+x−12)=0
⇒(x2−3x+2x−6)(x2+4x−3x−12)=0⇒(x2−3x+2x−6)(x2+4x−3x−12)=0
⇒[x(x−3)+2(x−3)][x(x+4)−3(x+4)]=0⇒[x(x−3)+2(x−3)][x(x+4)−3(x+4)]=0
⇒(x−3)(x+2)(x+4)(x−3)=0⇒(x−3)(x+2)(x+4)(x−3)=0
⇒⎡⎢⎣x−3=0x+2=0x+4=0⇒⎡⎢⎣x=3x=−2x=−4⇒[x−3=0x+2=0x+4=0⇒[x=3x=−2x=−4
c, x3−3x+2=0x3−3x+2=0
⇒x3+2x2−2x2−4x+x+2=0⇒x3+2x2−2x2−4x+x+2=0
⇒x2(x+2)−2x(x+2)+(x+2)=0⇒x2(x+2)−2x(x+2)+(x+2)=0
⇒(x+2)(x2−2x+1)=0⇒(x+2)(x2−2x+1)=0
⇒(x+2)(x−1)2=0⇒(x+2)(x−1)2=0
⇒[x+2=0(x−1)2=0⇒[x=−2x=1⇒[x+2=0(x−1)2=0⇒[x=−2x=1