Question

Tìm x biết 

a) x(x-5)(x+5)-(x-2)(x^2+2x+4)=3

b) 25x^2-2=0

c) (x+2)^2=(2x-1)^2

d) (x+2)^2-x^2+4=0

e) (x^2-2)^2+4(x-1)^2-4(x^2-2)(x-1)=0

 

 

Answer 1

\(x\left(x-5\right)\left(x+5\right)-\left(x-2\right)\left(x^2+2x+4\right)=3\)

<=> \(x\left(x^2-25\right)-\left(x^3+2x^2+4x-2x^2-4x-8\right)=3\)

<=> \(x^3-25x-x^3-2x^2-4x+2x^2+4x+8=3\)

<=> \(-25x+8=3\)

<=> \(-25x=-5\)

<=> \(x=\frac{1}{5}\)

\(25x^2-2=0\)

<=> \(\left(5x\right)^2=2\)

<=> \(\hept{\begin{cases}5x=\sqrt{2}\\5x=-\sqrt{2}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)

\(\left(x+2\right)^2=\left(2x-1\right)^2\)

<=> \(\hept{\begin{cases}x+2=2x-1\\x+2=-2x+1\end{cases}}\)

<=> \(\hept{\begin{cases}-x=-3\\3x=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}\)

\(\left(x+2\right)^2-x^2+4=0\)

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

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

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

<=> \(\left(x+2\right).4=0\)

<=> \(x+2=0\)

<=> \(x=-2\)

câu còn lại tương tự nha