Ta có :\(2x=5y\Rightarrow\dfrac{2}{5}=\dfrac{y}{x}\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}\) hay \(\dfrac{y}{2}=\dfrac{x}{5}\)

   Áp dụng tính chất tỉ lệ thức, ta có :

       \(\dfrac{y}{2}=\dfrac{x}{5}=\dfrac{y-x}{2-5}=\dfrac{12}{-3}=-4\)

\(\Rightarrow\dfrac{y}{2}=-4\Rightarrow y=2\times\left(-4\right)=-8\)

     \(\dfrac{x}{5}=-4\Rightarrow x=5\times\left(-4\right)=-20\)

                                    Vậy x= -4 ; y= -20

\(ĐKXĐ\left\{{}\begin{matrix}\dfrac{1}{-1+x}\ge0\\-1+x\ne0\end{matrix}\right.\)   ( Tử và mẫu cùng dấu )

   Mà 1 > 0 \(\Rightarrow-1+x>0\)

                  \(\Leftrightarrow\)          \(x>1\)

\(4x\left(3x+7\right)+2x\left(2-6x\right)=12x^2+28x+4x-12x^2=32x\)

mình nhìn nhầm rồi bạn ới, bài mk lm sai mất r nha 

\(a^2+2ab^2+4b^2=\left(a+2b^2\right)^2\)

\(8x^2-2x-3=8x^2+4x-6x-3=4x\left(2x+1\right)-3\left(2x+1\right)=\left(4x-3\right)\left(2x+1\right)\)

\(5x^2-5x-10=5x^2+5x-10x-10=5x\left(x+1\right)-10\left(x+1\right)=\left(5x-10\right)\left(x+1\right)\)

\(\left(2x-5\right).\left(4-3x\right)-\left(3x+11\right).\left(5-2x\right)-15\left(2x-5\right)\)

\(=\)\(\left(2x-5\right).\left(4-3x\right)+\left(3x+11\right).\left(2x-5\right)-15\left(2x-5\right)\)

\(=\left(2x-5\right).\left(4-3x+3x+11-15\right)\)

\(=\left(2x-5\right).0\)

\(=0\)