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Bài 3:
a) Ta có: \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4;-4\right\}\)
b) Ta có: \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)
c) Ta có: \(\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\)
d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
\(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\left(x+1\right)x=0\)
\(\orbr{\begin{cases}x+1=0\\x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)vậy.....
\(x\left(x-5\right)^2-4x+20=0\)
\(x\left(x-5\right)^2-4\left(x-5\right)=0\)
\(\left(x-5\right)\left[x\left(x-5\right)-4\right]=0\)
\(\left(x-5\right)\left(x^2-5x-4\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2-5x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-0,7015621187\end{cases}}}\)vậy.........
\(x\left(x+6\right)-7x-42=0\)
\(x\left(x+6\right)-7\left(x+6\right)=0\)
\(\left(x+6\right)\left(x-7\right)=0\)
\(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}}\) vậy....
\(x^3-5x^2+x-5=0\)
\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x^2=-1\Rightarrow x\in\Phi\end{cases}}}\)vậy........
\(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^3+10x\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)vậy..............
nhớ chọn mk nha
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-4=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=4\\x=-4\end{array}\right.\)
b) \(\left(2x-3\right)^2=\left(x-5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-3+x-5\right)\left(2x-3-x+5\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{8}{3}\end{array}\right.\)
c) \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
(x2 + x + 1)(6 - 2x) = 0
<=> 6 - 2x = 0 (do x2 + x + 1 > 0)
<=> 2x = 6
<=> x = 3
Vậy S = {3}
(8x - 4)(x2 + 2x + 2) = 0
<=> 8x - 4 = 0 (vì x2 + 2x + 2 > 0)
<=> 8x = 4
<=> x = 1/2
Vậy S = {1/2}
x3 - 7x + 6 = 0
<=> x3 - x - 6x + 6 = 0
<=> x(x2 - 1) - 6(x - 1) = 0
<=> x(x - 1)(x + 1) - 6(x - 1) = 0
<=> (x2 + x - 6)(x - 1) = 0
<=> (x2 + 3x - 2x - 6)(x - 1) = 0
<=> (x + 3)(x - 2)(x - 1) = 0
<=> x + 3 = 0
hoặc x - 2 = 0
hoặc x - 1 = 0
<=> x = -3
hoặc x = 2
hoặc x = 1
Vậy S = {-3; 1; 2}
x5 - 5x3 + 4x = 0
<=> x(x4 - 5x2 + 4) = 0
<=> x(x4 - x2 - 4x2 + 4) = 0
<=> x[x2(x2 - 1) - 4(x2 - 1)] = 0
<=> x(x - 2)(x + 2)(x - 1)(x + 1) = 0
<=> x = 0 hoặc x - 2 = 0 hoặc x + 2 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
<=> x = 0 hoặc x = 2 hoặc x = -2 hoặc x = 1 hoặc x = -1
Vậy S = {-2; -1; 0; 1; 2}
+ Ta có: \(\left(x^2+x+1\right).\left(6-2x\right)=0\)
- Ta lại có: \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
- Vì \(x^2+x+1>0\forall x\)mà \(\left(x^2+x+1\right).\left(6-2x\right)=0\)
\(\Rightarrow6-2x=0\Leftrightarrow-2x=-6\Leftrightarrow x=3\left(TM\right)\)
Vậy \(S=\left\{3\right\}\)
+ Ta có: \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)
- Ta lại có: \(x^2+2x+2=\left(x^2+2x+1\right)+1=\left(x+1\right)^2+1\ge1>0\forall x\)
- Vì \(x^2+2x+2>0\forall x\)mà \(\left(8x-4\right).\left(x^2+2x+2\right)=0\)
\(\Rightarrow8x-4=0\Leftrightarrow8x=4\Leftrightarrow x=\frac{1}{2}\left(TM\right)\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
+ Ta có: \(x^3-7x+6=0\)
\(\Leftrightarrow\left(x^3-x^2\right)+\left(x^2-x\right)+\left(6x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right).\left[\left(x^2-2x\right)+\left(3x-6\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right).\left(x-2\right).\left(x+3\right)=0\)
Vậy \(S=\left\{-3;1;2\right\}\)
+ Ta có: \(x^5-5x^3+4x=0\)
\(\Leftrightarrow x.\left[\left(x^4-x^2\right)-\left(4x^2-4\right)\right]=0\)
\(\Leftrightarrow x.\left[x^2.\left(x^2-1\right)-4.\left(x^2-1\right)\right]=0\)
\(\Leftrightarrow x.\left(x^2-1\right).\left(x^2-4\right)=0\)
\(\Leftrightarrow x=0\left(TM\right)\)
hoặc \(x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=\pm1\left(TM\right)\)
hoặc \(x^2-4=0\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\left(TM\right)\)
Vậy \(S=\left\{-2;-1;0;1;2\right\}\)
!!@@# ^_^ Chúc bạn hok tốt ^_^#@@!!
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......