Answer:

a) \(\frac{x}{x-1}+\frac{3}{x+1}-\frac{6x-4}{x^2-1}\) \(\left(ĐKXĐ:x\ne\hept{\begin{cases}x=1\\x=-1\end{cases}}\right)\)

\(=\frac{x}{x-1}+\frac{3}{x+1}-\frac{6x-4}{\left(x+1\right).(x-1)}\)

Mẫu thức chung: \(\left(x+1\right).\left(x-1\right)\)

\(B=\frac{x.\left(x+1\right)}{\left(x+1\right).\left(x-1\right)}+\frac{3.\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}-\frac{6x-4}{\left(x+1\right).\left(x-1\right)}\)

\(=\frac{x}{\left(x+1\right).\left(x-1\right)}+\frac{3x-3}{\left(x+1\right).\left(x-1\right)}-\frac{6x-4}{\left(x+1\right).\left(x-1\right)}\)

\(=\frac{x^2+x+3x-3-6x+4}{\left(x+1\right).\left(x-1\right)}\)

\(=\frac{x^2-2x+1}{\left(x+1\right).\left(x-1\right)}\)

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

b) \(B=5\)

\(\Rightarrow\frac{x-1}{x+1}=5\)

\(\Rightarrow x-1=5.\left(x+1\right)\)

\(\Rightarrow x-1=5x+5\)

\(\Rightarrow x-1-5x-5=0\)

\(\Rightarrow-4x-6=0\)

\(\Rightarrow-2.\left(2x+3\right)=0\)

\(\Rightarrow2x=-3\)

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

Answer:

a) ĐKXĐ: \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\)

\(A=\left(\frac{1}{x}+\frac{x}{x+1}\right):\frac{x}{x^2+x}\)

\(=\frac{x+1+x.x}{x.\left(x+1\right)}.\frac{x.\left(x+1\right)}{x}\)

\(=\frac{x^2+x+1}{x}\)

b) \(x=4\)

\(\Rightarrow A=\frac{4^2+4+1}{4}=\frac{21}{4}\)

c) \(A=3\)

\(\Rightarrow\frac{x^2+x+1}{x}=3\)

\(\Rightarrow x^2+x+1=3x\)

\(\Rightarrow x^2-2x+1=0\)

\(\Rightarrow\left(x-1\right)^2=0\)

\(\Rightarrow x=1\left(TMĐK\right)\)

b nhé anh :))))))))))))))))))))))))))))))

thanks

Answer:

\(A=\left(2x+5\right).\left(2x-5\right)-\left(4x-1\right).\left(x+3\right)\)

\(=\left(2x\right)^2-5^2-\left(4x^2+12x-x-3\right)\)

\(=4x^2-25-4x^2-12x+x+3\)

\(=-11x-22\)

\(B=\left(x-2\right)^2-6x.\left(x+5\right)+\left(3x-2\right).\left(2x+1\right)\)

\(=x^2-4x+4-6x^2-30x+6x^2+3x-4x-2\)

\(=x^2-35x+2\)