Question

giải phương trình

|x+1| = |x(x+1)|

Answer 1

\(\left|x+1\right|=\left|x\left(x+1\right)\right|\)

+) Xét \(x\ge-1\) ta có:
\(x+1=x\left(x+1\right)\)

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

\(\Rightarrow x^2=1\)

\(\Rightarrow x=\pm1\) ( t/m )

+) Xét \(x< -1\) ta có:
\(-\left(x+1\right)=x\left(x+1\right)\)

\(\Rightarrow x\left(x+1\right)+\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)^2=0\)

\(\Rightarrow x+1=0\Rightarrow x=-1\) ( loại )

Vậy \(x\in\left\{1;-1\right\}\)

Answer 2

|x+1|= |x(x+1)|

<=> x+1 = x(x+1) hoac x+1 = - x(x+1)

<=> x+1 = x²+x hoac x+1 = - x²-x

<=> x - x² - x = -1 hoac x + x² + x + 1 = 0

<=> x²=1 hoac x²+2x+1=0

<=> x=1 hoac x= -1 hoac (x+1)²=0

<=> x=1 hoac x= -1

S= { -1 ; 1 }