Question

Cho \(A=\frac{x-1}{x+3}\) và    \(B=\frac{1}{x+3}-\frac{x}{x-1}-\frac{4x}{x^2+2x-3}\)  \(\left(x\ge0;x\ne1\right)\)

a, Rút gọn B

b,tìm x để \(\frac{A-1}{B}\le\frac{1}{2}\)

Answer 1

chịu . Tui mới lớp 5

Answer 2

a) Ta có: 

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

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

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

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

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

b) xem lại đề