\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)( ĐKXĐ : \(x\ne\pm2\))
\(=\frac{5\left(x+2\right)}{2\left(2x-4\right)}\cdot\frac{-\left(2x-4\right)}{x+2}\)
\(=\frac{-5\left(x+2\right)\left(2x-4\right)}{2\left(2x-4\right)\left(x+2\right)}\)
\(=-\frac{5}{2}\)
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}\)( ĐKXĐ : \(x\ne-5;x\ne6\))
\(=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}\cdot\frac{3}{-\left(x-6\right)}\)
\(=\frac{3\left(x-6\right)\left(x+6\right)}{-2\left(x+5\right)\left(x-6\right)}\)
\(=\frac{3\left(x+6\right)}{-2\left(x+5\right)}=\frac{3x+18}{-2x-10}=-\frac{3x+18}{2x+10}\)
a)
Điều kiện : \(\hept{\begin{cases}4x-8\ne0\\x+2\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}\)
\(=\frac{5\left(x+2\right)}{-2\left(4-2x\right)}\cdot\frac{4-2x}{x+2}\)
\(=\frac{-5}{2}\)
b)
Điều kiện : \(\hept{\begin{cases}2x+10\ne0\\6-x\ne0\end{cases}}\)
\(\hept{\begin{cases}x\ne-5\\x\ne6\end{cases}}\)
\(=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}\)
\(=\frac{-6\left(x+6\right)\cdot3}{2x+10}\)
\(=\frac{-9\left(x+6\right)}{x+5}\)
\(=\frac{-9x-54}{x+5}\)
\(=\frac{-9\left(x+5\right)-9}{x+5}\)
\(=-9-\frac{9}{x+5}\)
\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}\cdot\frac{-2\left(x-2\right)}{x+2}=\frac{-10\left(x+2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}=-\frac{5}{2}\)
b) \(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{3\left(x^2-36\right)}{2\left(x+5\right)\left(6-x\right)}=\frac{3\left(x-6\right)\left(x+6\right)}{-2\left(x-6\right)\left(x+5\right)}=\frac{3\left(x+6\right)}{-2\left(x+5\right)}=\frac{3x+18}{-2x-10}\)
P/S : K chắc :<
Xin lỗi mình nhầm câu B
Câu B bạn tham khảo 2 bạn ở dưới nha
mình thêm nhầm 6 vào chỗ \(\frac{-6\left(x+6\right)\cdot3}{2x+10}\)
Bỏ số 6 trước ngoặc nha
Xin lỗi vì sự nhầm lẫn
