Question

Tìm x :(x+4)^2+(1-x)(1+x)=7

Answer 1

Lời giải:

$(x+4)^2+(1-x)(1+x)=7$

$\Leftrightarrow x^2+8x+16+(1-x^2)=7$

$\Leftrightarrow 8x+17=7$

$\Leftrightarrow 8x=-10$

$\Leftrightarrow x=\frac{-10}{8}=\frac{-5}{4}$

Answer 2

Ta có: \(\left(x+4\right)^2+\left(1-x\right)\left(1+x\right)=7\)

\(\Leftrightarrow x^2+8x+16+1-x^2=7\)

\(\Leftrightarrow8x=-10\)

hay \(x=-\dfrac{5}{4}\)