Question

Cho biểu thức: M=\(\frac{2x}{1-x^2}\)(\(\frac{1}{x^2+2x+1}-\frac{1}{x^2-1}\)) (Với x\(\ne\)+-1)

a)Rút gọn M

b)Tìm giá trị nhỏ nhất của M

Answer 1

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

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

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

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

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

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