Giải câu 3 trang 28 sách phát triển năng lực toán 9 tập 1

a, Với $x\geq 0,x\neq 1$ ta có:

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

b, Với $x\geq 0$ ta có:

$\frac{(2+\sqrt{3x})^{2}-(\sqrt{3x}+1)^{2}}{2\sqrt{3x}+3}$ = $\frac{4+4\sqrt{3x}+3x-(3x+2\sqrt{3x}+1)}{2\sqrt{3x}+3}$ = $\frac{2\sqrt{3x}+3}{2\sqrt{3x}+3}=1$

c, Với $x\geq 0,x\neq 4$ ta có:

   $\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}$ = $\frac{\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}-\frac{2.(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{\sqrt{x}}{4-x}$ 

= $\frac{\sqrt{x}-2-2.(\sqrt{x}+2)}{x-4}-\frac{\sqrt{x}}{4-x}$ = $\frac{-\sqrt{x}-6}{x-4}-\frac{\sqrt{x}}{4-x}$ =$\frac{\sqrt{x}+6}{4-x}-\frac{\sqrt{x}}{4-x}$ = $\frac{\sqrt{x}+6-\sqrt{x}}{4-x}=\frac{6}{4-x}$

d, Với $x\geq \frac{1}{2}$ ta có:

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

e, Với $x\geq 0$ ta có:

$\frac{\sqrt{a}}{\sqrt{a+1}-\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a+1}+\sqrt{a}}$ = $\frac{\sqrt{a}.(\sqrt{a+1}+\sqrt{a})}{(\sqrt{a+1}-\sqrt{a})(\sqrt{a+1}+\sqrt{a})}+\frac{\sqrt{a}.(\sqrt{a+1}-\sqrt{a})}{(\sqrt{a+1}+\sqrt{a}).(\sqrt{a+1}-\sqrt{a})}$

= $\frac{\sqrt{a}.(\sqrt{a+1}+\sqrt{a}+\sqrt{a+1}-\sqrt{a})}{a+1-a}$ = $\sqrt{a}.2\sqrt{a+1}=2\sqrt{a(a+1)}$

f, $\frac{1}{x+\sqrt{1+x^{2}}}+\frac{1}{x-\sqrt{1+x^{2}}}+2x$ = $\frac{1.(x-\sqrt{1+x^{2}})}{(x+\sqrt{1+x^{2}})(x-\sqrt{1+x^{2}})}+\frac{1.(x+\sqrt{1+x^{2}})}{(x-\sqrt{1+x^{2}})(x+\sqrt{1+x^{2}})}+2x$

= $\frac{(x-\sqrt{1+x^{2}})+(x+\sqrt{1+x^{2}})}{x^{2}-(1+x^{2})}+2x$ = $\frac{2x}{-1}+2x = 0$