Question

tìm x

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

Answer 1

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

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

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

\(x+2=0\)

\(x=-2\)

Answer 2

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

\(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

Vậy.............

Answer 3

(x+2)²-(x-2)(x+2)=0

=> (x+2)[(x+2)-(x-2)]=0

=> (x+2)(x+2-x+2)=0

=>(x+2)+4=0

=> x+4=-4

=> x=-8

vậy ...

Answer 4

\(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\)

Vậy \(x=-2\)