b) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

\(=\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}+\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}}\)

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

\(=\dfrac{3+3}{2}=\dfrac{6}{2}=3\)

Lời giải:
1/

\(=\frac{3.\sqrt{5}-\sqrt{5}.\sqrt{15}}{\sqrt{15}-3}=\frac{-\sqrt{5}(\sqrt{15}-3)}{\sqrt{15}-3}=-\sqrt{5}\)

2/

\(=\frac{\sqrt{2+2\sqrt{2.3}+3}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{(\sqrt{2}+\sqrt{3})^2}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}}=1\)

3/

\(=\frac{2^2+2.2\sqrt{3}+3}{2+\sqrt{3}}=\frac{(2+\sqrt{3})^2}{2+\sqrt{3}}=2+\sqrt{3}\)

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)

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

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

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

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

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

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...

\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)

\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)

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

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

\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)

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

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

a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)

\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)

b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)

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

c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)

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

d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)

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

Bài 1:

a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)

\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)

\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)

\(=\dfrac{5}{2}\sqrt{5}\)

c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)

\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)

Bài 2:

e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)

Ta có: \(\sqrt{6-x}=3x-4\)

\(\Leftrightarrow6-x=\left(3x-4\right)^2\)

\(\Leftrightarrow9x^2-24x+16+6-x=0\)

\(\Leftrightarrow9x^2-25x+22=0\)

\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)

Vậy: Phương trình vô nghiệm

Bài 1: 

\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}.4\sqrt{3}-\sqrt{3}+5.\dfrac{2\sqrt{3}}{3}=2\sqrt{3}-\sqrt{3}+\dfrac{10\sqrt{3}}{3}=\dfrac{3\sqrt{3}+10\sqrt{3}}{3}=\dfrac{13\sqrt{3}}{3}\)

\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}=\sqrt{\dfrac{\left(3-\sqrt{5}\right)^2}{9-5}}-\sqrt{\dfrac{\left(3+\sqrt{5}\right)^2}{9-5}}=\dfrac{3-\sqrt{5}}{2}-\dfrac{3+\sqrt{5}}{2}=-\sqrt{5}\)

\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\sqrt{\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}+\sqrt{\left(\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}\right)^2}-\sqrt{2}=\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{1}{2}}+\dfrac{3}{\sqrt{2}}-\sqrt{\dfrac{5}{2}}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2}\)

Bài 2: 

Ta có: G-1

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

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

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le0\forall x\) thỏa mãn ĐKXĐ

hay \(G\le1\)

Lời giải:
a.

\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)

b.

\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)

\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)

\(=10\)

c.

\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)

\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)

d.

\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)

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

Lời giải:

$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$

\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)

Do đó:

\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)

\(A=\frac{\text{TS}}{\text{MS}}=1\)

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{2}-\dfrac{7\left(2+\sqrt{3}\right)}{4-3}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\left(\sqrt{5}+\sqrt{3}\right)-14-7\sqrt{3}+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}+4\sqrt{3}-6\sqrt{3}+4\sqrt{2}-11\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)

\(M=\dfrac{8\left(\sqrt{5}+\sqrt{3}\right)}{5-3}+\dfrac{7\left(\sqrt{3}+2\right)}{3-4}+\dfrac{4\left(\sqrt{2}+1\right)}{2-1}+\dfrac{\sqrt{15}\left(\sqrt{3}-1\right)}{\sqrt{15}}\)

\(=4\sqrt{5}+4\sqrt{3}-7\sqrt{3}-14+4\sqrt{2}+4+\sqrt{3}-1\)

\(=4\sqrt{5}-2\sqrt{3}+4\sqrt{2}-11\)