\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)

\(=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

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

\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)

\(=\dfrac{3}{2}\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)

\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

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

\(C=2-\sqrt{3}+3+\sqrt{3}=5\)

a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)

\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)

=3

Lời giải:
ĐKXĐ: $x>0$

a. \(P=\frac{x-1}{\sqrt{x}}:\left[\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)}+\frac{1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}\right]\)

\(=\frac{x-1}{\sqrt{x}}:\frac{x-1+1-\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)}=\frac{x-1}{\sqrt{x}}:\frac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}+1)}=\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}}.\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

b.

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

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

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

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

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

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

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)

\(=\dfrac{5-3\sqrt{5}+10+6\sqrt{5}}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}-\dfrac{2\sqrt{10}+2}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{15+3\sqrt{5}}{5-9}-\left(2\sqrt{10}+2\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =-2\sqrt{30}-4\sqrt{5}-2\sqrt{3}-2\sqrt{2}-\dfrac{15+3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-16\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15-3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-19\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15}{4}\)

`a)(5sqrt2-2sqrt5)/(sqrt5-sqrt2)+6/(2-sqrt{10})`

`=(sqrt{10}(sqrt5-sqrt2))/(sqrt5-sqrt2)+(6(2+sqrt{10}))/(4-10)`

`=sqrt{10}-(2+sqrt{10})`

`=-2`

`b)6/(sqrt5-1)+7/(1-sqrt3)-2/(sqrt3-sqrt5)`

`=(6(sqrt5+1))/(5-1)+(7(1+sqrt3))/(1-3)-(2(sqrt3+sqrt5))/(3-5)`

`=(6(sqrt5+1))/4-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3)/2-(7+7sqrt3)/2+sqrt3+sqrt5`

`=(3sqrt5+3-7-7sqrt3+2sqrt3+2sqrt5)/2`

`=(5sqrt5-5sqrt3-4)/2`

\(A=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(đk:a>0,a\ne1\right)\)

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+2}=\dfrac{1}{\sqrt{a}}.\dfrac{\sqrt{a}-2}{1}=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)

Để A nguyên

\(\Leftrightarrow A=\dfrac{\sqrt{a}-2}{\sqrt{a}}=1-\dfrac{2}{\sqrt{a}}\in Z\)

Do \(\sqrt{a}>0,\sqrt{a}\ne1\)

\(\Leftrightarrow\sqrt{a}\inƯ\left(2\right)=\left\{2\right\}\)

\(\Leftrightarrow a=4\)

\(a,=\sqrt{3}+4\sqrt{3}+20\sqrt{3}-10\sqrt{3}=15\sqrt{3}\\ b,=4\sqrt{5}+\sqrt{5}-1-\dfrac{20\left(\sqrt{5}-1\right)}{4}\\ =5\sqrt{5}-1-5\sqrt{5}+5=4\\ c,=\dfrac{6\sqrt{13}+6+6\sqrt{13}-6}{\left(\sqrt{13}-1\right)\left(\sqrt{13}+1\right)}=\dfrac{12\sqrt{13}}{12}=\sqrt{13}\\ d,=\left(\sin^238^0+\cos^238^0\right)+\left(\tan67^0-\tan67^0\right)=1+0=1\)