Bài 2

\(P=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{4-\sqrt{12}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)

\(=\frac{\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)

\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\left(\sqrt{3}+1\right)}\)

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

\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}+1\right)}=1\)

Vậy P là một số nguyên

\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)

\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{-\sqrt{x}+5}\)

\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}}{5-\sqrt{x}}\)

\(=-\dfrac{x}{5-\sqrt{x}}\)

\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

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

\(=\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\)

\(=2+\sqrt{3}+2-\sqrt{3}=4\)

\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)

NX \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)

\(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)

\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)=\(\frac{a^4+2a^3+2a^2+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)

\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)=\(\left[\frac{a^2+a+1}{a\left(a+1\right)}\right]^2\)suy ra A=\(\frac{a^2+a+1}{a\left(a+1\right)}\)

                                                                                                =\(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)

ap dung vao bai ta co =\(\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\)

=\(2011+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)= \(2011+\frac{1}{2}-\frac{1}{2013}=2011,499503\)

`a)P=(x^2+sqrtx)/(x-sqrtx+1)-(2x+sqrtx)/sqrtx`

`P=(sqrtx(sqrtx+1)(x-sqrtx+1))/(x-sqrtx+1)-(sqrtx(2sqrtx+1))/sqrtx`

`P=x+sqrtx-2sqrtx-1`

`P=x-sqrtx-1`

a: Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)

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

\(=x-\sqrt{x}-1\)

Ta có: M= \(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)= \(\frac{\left(1+2x\right)\left(1-\sqrt{1+2x}\right)+\left(1-2x\right)\left(1+\sqrt{1+2x}\right)}{1-\left(1-2x\right)}\)=\(\frac{1-\sqrt{1+2x}+2x-2x\sqrt{1+2x}+1+\sqrt{1+2x}-2x-2x\sqrt{1+2x}}{2x}\)

=\(\frac{2}{2x}=\frac{1}{x}\)

Với x=\(\frac{\sqrt{3}}{4}\)=> M=\(\frac{4}{\sqrt{3}}\)

bài này dài phết @@

M = \(\dfrac{4}{\sqrt{3}}\)