\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{2}\right)\)\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{3x}{3\left(4-x^2\right)}-\dfrac{1}{x+2}\right)\)\(=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{2}{x-2}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2x+4-3x-x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=1+\dfrac{1}{x+2}:\left(\dfrac{-2x+6}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=1+\dfrac{1}{x+2}.\dfrac{\left(x-2\right)\left(x+2\right)}{-2x+6}\)

\(=1+\dfrac{x-2}{-2x+6}\)

\(=\dfrac{-2x+6+x-2}{-2x+6}=\dfrac{4-x}{-2\left(x-3\right)}\)

\(\left(4x-5\right)^2-\left(3-6x\right)^2=0\)

\(\Leftrightarrow\left(4x-5-3+6x\right)\left(4x-5+3-6x\right)=0\)

\(\Leftrightarrow\left(10x-8\right)\left(-2x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10x-8=0\\-2x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}10x=8\\-2x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-1\end{matrix}\right.\)

híu bé ơi ko sợ cj ném đá hả

you nhờ thầy ik mk cx nhờ thầy mk đổi đc nà

Đề bài đúng chứ bạn

lại có trào lưu ms-_-