\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)

\(=\frac{4}{x-1}+\frac{-2}{x-1}-\frac{x}{x-1}\)

\(=\frac{2-x}{x-1}\)

ĐKXĐ: \(x\ne1\)

\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)

\(=\frac{4}{x-1}+\frac{2}{x-1}-\frac{x}{x-1}\)

\(=\frac{4+2-x}{x-1}\)

\(=\frac{6-x}{x-1}\)

a) \(\frac{6xy+4y}{4x^2y^2}+\frac{2xy-4y}{4x^2y^2}\)

\(=\frac{6xy+4y+2xy-4y}{4x^2y^2}\)

\(=\frac{8xy}{4x^2y^2}\)

\(=\frac{2}{xy}\)

b) \(\frac{5}{x+3}-\frac{3}{x-3}+\frac{30}{x^2-9}\)

\(=\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{30}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{5x-15-3x-9+30}{\left(x+3\right)\left(x-3\right)}\)

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

\(=\frac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{2}{x-3}\)

c) \(\frac{2x+8}{\left(x+2\right)^2}:\frac{x+4}{x+2}\)

\(=\frac{2\left(x+4\right)}{\left(x+2\right)^2}\cdot\frac{x+2}{x+4}\)

\(=\frac{2\left(x+4\right)\left(x+2\right)}{\left(x+2\right)\left(x+2\right)\left(x+4\right)}\)

\(=\frac{2}{x+2}\)

\(\frac{3x-6}{x-2}=\frac{3\left(x-2\right)}{x-2}=3\)

ĐKXĐ: \(x\ne2\)

\(\frac{3x-6}{x-2}\)

\(=\frac{3.\left(x-2\right)}{x-2}\)

\(=3\)

đáp án:

1+1=2

so ez

1 + 1 = 2

Hok tốt

(-5)+(-11)

\(\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+36=0\)

\(\left[\left(x-1\right)\left(x-8\right)\right]\left[\left(x-4\right)\left(x-5\right)\right]+36=0\)

\(\left(x^2-9x+8\right)\left(x^2-9x+20\right)+36=0\)

Đặt \(a=x^2-9x+14\)ta có :

\(\left(a-6\right)\left(a+6\right)+36=0\)

\(a^2-6^2+36=0\)

\(a^2=0\)

Thay \(a=x^2-9x+14\)ta có :

\(\left(x^2-9x+14\right)^2=0\)

\(\Leftrightarrow x^2-9x+14=0\)

\(\Leftrightarrow x^2-2x-7x+14=0\)

\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=7\end{cases}}}\)

Vậy,...........

dấu ngang ngang là phần nha