Question

Tìm x biết  :

a.  x³-25x =0

b.  4x(x+1)=8(x+1)

c.   4-x=4(x-4)²

Answer 1

`x^3 -25x=0`

``x.(x-5).(x+5)=0`

`=>`\(\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

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`4x.(x+1)=8.(x+1)`

`4x.(x+1)-8.(x+1)=0`

`(x+1).(x-2)=0`

`=>`\(\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

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`4-x=4.(x-4)^2`

`4-x-4.(x-4)^2 = 0`

`4-x-4.(x^2 - 8x + 16)=0`

`4-x-4x^2 +32x - 64=0`

`-4x^2 - 31x -60=0`

`4x^2 - 31x +60=0`

`(4+x).(15-4x)=0`

`=>`\(\left[{}\begin{matrix}4+x=0\\15-4x=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=-4\\x=\dfrac{15}{4}\end{matrix}\right.\)

Answer 2

a) \(x^3-25x=0\)

\(x\left(x^2-25\right)\)

=> \(\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Answer 3

b) \(4x\left(x+1\right)=8\left(x+1\right)\)

Ta thấy (x+1) chung và 2 biểu thức trên bằng nhau

=> 4x=8

=> x=2

Answer 4

a)

\(< =>x\left(x^2-25\right)=0< =>\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

b)

\(< =>4x\left(x+1\right)-8\left(x+1\right)=0< =>\left(4x-8\right)\left(x+1\right)=0< =>\left[{}\begin{matrix}4x-8=0\\x+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)c)

\(< =>4-x-4\left(x-4\right)^2=0< =>\left(x-4\right)\left[1+4\left(x-4\right)\right]=0< =>\left(x-4\right)\left(1+4x-16\right)=0< =>\left(x-4\right)\left(-15+4x\right)=0< =>\left[{}\begin{matrix}x-4=0\\-15+4x=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=4\\x=\dfrac{15}{4}\end{matrix}\right.\)

Answer 5

a) \(x^3-25x=0\)

\(\Leftrightarrow x\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)

`=>`\(x\in\left\{0;-5;5\right\}\)

b) \(4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x^2+4x=8x+8\)

`<=>`\(4x^2+4x-8x=8\)

`<=>`\(4x^2-4x=8\)

`<=>`\(4x^2-4x-8=0\)

`<=>`\(4\left(x-2\right)\left(x+1\right)=0\)

`<=>`\(\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\) 

`=>`\(x\in\left\{2;-1\right\}\)

c) \(4-x=4\left(x-4\right)^2\)

`<=>`\(4-x=4x^2-32x+64\)

\(\Leftrightarrow4-x-4x^2+32x-64=0\)

\(\Leftrightarrow4+31x-4x^2-64=0\)

\(\Leftrightarrow-60+31x-4x^2=0\)

\(\Leftrightarrow4x^2-16x-15x+60=0\)

\(\Leftrightarrow4x\left(x-4\right)-15\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x-15=0\end{matrix}\right.\)

`=>`\(x\in\left\{\dfrac{15}{4};4\right\}\)