bài 6: rút gọn

giúp tớ với mn ơiii

Giải thích các bước giải:

e.Ta có:

$E=\dfrac{\sqrt{x}-1}{x+\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}$

$\to E=\dfrac{\sqrt{x}- 1}{\sqrt{x}(\sqrt{x}+1)}+\dfrac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$\to E=\dfrac{(\sqrt{x}-1)^2+(\sqrt{x})^2}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}$

$\to E=\dfrac{x-2\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}$

$\to E=\dfrac{2x-2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}$

f.Ta có:

$F=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}-\dfrac{\sqrt{x}}{x-4\sqrt{x}+4}$

$\to F=\dfrac{\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-2)}-\dfrac{\sqrt{x}}{(\sqrt{x}-2)^2}$

$\to F=\dfrac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)^2}-\dfrac{x}{\sqrt{x}(\sqrt{x}-2)^2}$

$\to F=\dfrac{(\sqrt{x}+2)(\sqrt{x}-2)-x}{\sqrt{x}(\sqrt{x}-2)^2}$

$\to F=\dfrac{x-4-x}{\sqrt{x}(\sqrt{x}-2)^2}$

$\to F=\dfrac{-4}{\sqrt{x}(\sqrt{x}-2)^2}$

g.Ta có:

$G=\dfrac{\sqrt{x}-3}{x+3\sqrt{x}}-\dfrac{\sqrt{x}+3}{x-3\sqrt{x}}$

$\to G =\dfrac{(\sqrt{x}-3)^2-(\sqrt{x}+3)^2}{\sqrt{x}(\sqrt{x}+3)(\sqrt{x}-3)}$

$\to G=\dfrac{-12\sqrt{x}}{\sqrt{x}(x-9)}$

$\to G=\dfrac{-12}{x-9}$

h.Ta có:

$H=\dfrac{\sqrt{x}+1}{x-\sqrt{x}}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}}$

$\to H=\dfrac{(\sqrt{x}+1)^2-(\sqrt{x}-1)^2}{\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}$

$\to H=\dfrac{4\sqrt{x}}{\sqrt{x}(x-1)}$

$\to H=\dfrac4{x-1}$