Question

tìm x

a, (x+3)^2 -x^2 +9 = 0

b, (x-2) (x^2+2)+(2-x)3x = 0

Answer 1

a, \(\left(x+3\right)^2-x^2+9=0\)

\(\Leftrightarrow x^2+6x+9-x^2+9=0\)

\(\Leftrightarrow6x+18=0\)

\(\Leftrightarrow6x=-18\)

\(\Leftrightarrow x=-3\)

Vậy x = -3.

Answer 2

b, \(\left(x-2\right)\left(x^2+2\right)+\left(2-x\right)3x=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2\right)+\left[-\left(x-2\right)\right]3x=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2\right)-\left(x-2\right)3x=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+3x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x^2+x\right)+\left(2x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)=0\)

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

Vậy \(x\in\left\{2;-1;-2\right\}\)