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\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ 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 ~ )
\(j,3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy...............................
\(m,3x^2+4x-4=0\)
\(\Leftrightarrow3x^2+6x-2x-4=0\)
\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)
a: \(6x^4+25x^3+12x^2-25x+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(3x-1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{-2;-3;\dfrac{1}{3};\dfrac{1}{2}\right\}\)
b: \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^2+x^3+x+x^2+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
=>x+1=0
hay x=-1
c: \(x^2\left(x^2+2\right)-x^2-2=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
b. sửa đề
\(6x^4+25x^3+12x-25x^2+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-3\\x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy........
Bài 1 : Giải phương trình
a) (x + 3)4 + (x + 5)4 = 16
Đặt : x + 3 = t
=> x + 5 = x + 3 + 2 = t + 2
Thay x + 3 = t và x + 5 = t + 2 vào phương trình, ta có :
t4 + (t + 2)4 = 16
<=> 2t4 + 8t3 + 24t2 + 32t + 16 = 16
<=> 2(t4 + 4t3 + 12t2 + 16t) = 0
<=> t4 + 4t3 + 12t2 + 16t = 0
<=> (t + 2) . t . (t2 + 2y + 4) = 0
TH1 : t = 0
TH2 : t + 2 = 0 <=> t = -2
TH3 : t2 + 2y + 4 = 0 (vô nghiệm => loại)
Nên t = 0 hoặc t = -2
hay x + 3 = -2 hoặc x + 3 = 0
<=> x = -5 hoặc x = -3
\(S=\left\{-5;-3\right\}\)
b) 6x4 + 25x3 + 12x2 - 25x + 6 = 0
<=> 6x4 + 12x3 + 13x3 + 26x2 - 14x2 - 28x + 3x + 6 = 0
<=> 6x3 (x + 2) + 13x2 (x + 2) - 14x (x + 2) + 3(x + 2) = 0
<=> (x + 2)(6x3 + 13x2 - 14x + 3) = 0
<=> (x + 2)(6x3 + 18x2 - 5x2 - 15x + x + 3) = 0
\(\Leftrightarrow\left(x+2\right)[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)]=0\)
<=> (x + 2)(x + 3) (6x2 - 5x + 1) = 0
<=> (x + 2)(x + 3)(2x - 1)(3x - 1) = 0
TH1 : x + 2 = 0 <=> x = -2
TH2 : x + 3 = 0 <=> x = -3
TH3 : 2x - 1 = 0 <=> 2x = 1 <=> x = \(\dfrac{1}{2}\)
TH4 : 3x - 1 = 0 <=> 3x = 1 <=> 3x = \(\dfrac{1}{3}\)
\(S=\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)
a)\(6x^2+5x-6=0\)
\(\Leftrightarrow6x^2-4x+9x-6=0\)
\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
b)\(6x^2-13x+6=0\)
\(\Leftrightarrow6x^2-4x-9x+6=0\)
\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
c)\(10x^2-13x-3=0\)
\(\Leftrightarrow10x^2-15x+2x-3=0\)
\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)
d)\(20x^2+19x-3=0\)
\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)
\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)
e)\(3x^2-x+6=0\)
\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)
Suy ra vô nghiệm
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 \(-2\)
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
a) Gần giống cho nó giống luôn.
cần thêm (-x^3+2x^2-x) là giống
\(\left(x-1\right)^4+x^3-2x^2+x=\left(x-1\right)^4+x\left(x^2-2x+1\right)=\left(x-1\right)^4+x\left(x-1\right)^2\)
\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=0\\\left(x-1\right)^2+x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
Nghiệm duy nhất: x=1
câu d