Question

Cho hai biểu thức

A=\(\frac{x^2+7}{x}\) và B=\(\frac{x}{x+3}\)+\(\frac{2x-1}{x-3}\)-\(\frac{2x^2-x-3}{x^2-9}\)

a) Rút gọn B

b) Cho x>0 tìm GTNN của M= A+\(\frac{1}{B}\)

Answer 1

a, B = \(\frac{x}{x+3}+\frac{2x+1}{x-3}-\frac{2x^2-x-3}{x^2-9}\)

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

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

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

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