Question

|8-x|=x^2+x

Answer 1

+) Nếu 8 - x ≥ 0 <=> x ≤8 thì:

8 -x = x² + x

<=> -x² - 2x + 8 = 0

<=> -x² - 4x + 2x + 8 = 0

<=> -x(x + 4) + 2(x + 4) = 0

<=> (x+4)(2-x) = 0

<=> \(\left[{}\begin{matrix}x+4=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\) (t/m)

+) Nếu 8 - x < 0 <=> x > 8 thì:

8 - x = -x² - x

<=> x² + 8 = 0

<=> x² = -8 (vô lí)

Vậy pt có 2 nghiệm x = -4 hoặc x = 2

Answer 2

|8-x|>0➝ x>8

TA có \(8-x=x^2+x\)

\(x^2+2x+1=9\)

\(\left(x+1\right)^2=9\)

➩ x+1 =3➩x=2

➩x+1=-3➩x=-4