Question

giải pt (x+1)(x+2)=(2-x)(x+2)

Answer 1

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

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

\(\Leftrightarrow\left(x+2\right)\left(x+1-2+x\right)=0\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{2}\)

Vậy tập nghiệm của phương trình là S = { -2 ; 1/2 } 

Answer 2

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

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

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

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

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

Vậy: \(S=\left\{-2;\dfrac{1}{2}\right\}\)

Answer 3

pt <=> ( x + 1 )( x + 2 ) - ( 2 - x )( x + 2 ) = 0

<=> ( x + 2 )( 2x - 1 ) = 0

<=> x = -2 hoặc x = 1/2

Vậy pt có tập nghiệm S = { -2 ; 1/2 }