Question

tính giá trị của:

C= \(\frac{x+1}{x-3}\)-\(\frac{1-x}{x+3}\)+\(\frac{2x\left(x-1\right)}{9-x^2}\) tại x= 5

các bạn xin hãy giúp mik

Answer 1

C= \(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\)

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

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

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