Câu 1:
\(y^2-3y=0\Leftrightarrow y\left(y-3\right)=0\Leftrightarrow\left[{}\begin{matrix}y=0\\y=3\end{matrix}\right.\)
Câu 2:
\(a,=\left|\dfrac{4x+16-4x+16}{\left(x-4\right)\left(x+4\right)}\right|\cdot\dfrac{\left(x+4\right)^2}{32}=\dfrac{32}{\left|\left(x-4\right)\left(x+4\right)\right|}\cdot\dfrac{\left(x+4\right)^2}{32}=\dfrac{\left(x+4\right)^2}{\left|\left(x-4\right)\left(x+4\right)\right|}\)
Với \(x\le-4;x\ge4\Leftrightarrow BT=\dfrac{\left(x+4\right)^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{x+4}{x-4}\)
Với \(-4< x< 4\Leftrightarrow BT=\dfrac{\left(x+4\right)^2}{\left(4-x\right)\left(x+4\right)}=\dfrac{x+4}{4-x}\)
\(b,=\dfrac{5x}{3}\\ c,=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\\ d,=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+2\right)\left(x+3\right)}=\dfrac{x+4}{x+2}\)
Câu 3:
\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}=2\left(-\dfrac{3}{2}\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\\ B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{5}{15}=\dfrac{1}{3}\)
cảm ơn mọi người đã chỉ em đang cần gấp ạ câu 2,3,4,5 làm mỗi 4 câu đó thôi ạ
