Question

Answer 1

a: ĐKXĐ: x∉{1;-1}

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

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

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

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

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

b: B=5

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

=>5(x+1)=x-1

=>5x+5=x-1

=>4x=-6

=>\(x=-\frac64=-\frac32\) (Nhận)