Question

bài 1 thịc hiện phép tính

a) \(\frac{1}{1-x}\)+\(\frac{2x}{x^2-1}\)

b) \(\frac{1}{xy-x^{2^{ }}}\)-\(\frac{1}{y^2-xy}\)

c)\(\frac{4x+1}{2}\)-\(\frac{3x-2}{3}\)

d) \(\frac{x+3}{x}\)-\(\frac{x}{x-3}+\frac{9}{x^2-3x}\)

e)\(\frac{x+3}{x^2-1}-\frac{1}{x^2+x}\)

g) \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)

h)\(\frac{42a^2-3a+5}{a^3-1}-\frac{1-2a}{a^2+a+1}-\frac{6}{a-1}\)

f)\(\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\)

k)\(\frac{52x^2-y^2}{xy}-\frac{3x-2y}{y}\)

Answer 1

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

Answer 2

c) \(\frac{4x+1}{2}-\frac{3x-2}{3}\\ =\frac{3.\left(4x+1\right)-2.\left(3x-2\right)}{6}=\frac{6x+7}{6}\)

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