a. ta có:
M =\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}\)- \(\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)+\(\dfrac{x+1}{\sqrt{x}}\)(ĐKXĐ: x>0 và x≠1)
= \(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)-\(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)+\(\dfrac{x+1}{\sqrt{x}}\)
= \(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)-\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\)+\(\dfrac{x+1}{\sqrt{x}}\)
=\(\dfrac{x+\sqrt{x}+1-1+\sqrt{x}-1+x+1}{\sqrt{x}}\)
=\(\dfrac{2x+2\sqrt{x}}{\sqrt{x}}\)=\(\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)=\(2\left(\sqrt{x}+1\right)\)
b. để M=\(\dfrac{9}{2}\) thì \(2\left(\sqrt{x}+1\right)\)= \(\dfrac{9}{2}\) ⇔ \(\sqrt{x}+1\)=\(\dfrac{9}{4}\)
⇔ \(\sqrt{x}\)=\(\dfrac{5}{4}\) ⇔ x = \(\dfrac{25}{16}\)(TMĐKXĐ)
Vậy với x=\(\dfrac{25}{16}\) thì M=\(\dfrac{9}{2}\)
mk ko làm ra đc câu c. bạn thông cảm nha.....
