bài 1: 

a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)

\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)

\(=-33\sqrt{2}\)

b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)

\(=10-2\sqrt{21}+14\sqrt{21}\)

\(=12\sqrt{21}+10\)

Bài 2: 

a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)

\(\Leftrightarrow\left|2x+3\right|=8\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)

b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)

\(\Leftrightarrow4\sqrt{x}=8\)

hay x=4

c: Ta có: \(\sqrt{9x-9}+1=13\)

\(\Leftrightarrow3\sqrt{x-1}=12\)

\(\Leftrightarrow x-1=16\)

hay x=17

\(A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}\)

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

\(B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}\)

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

\(C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}\)

\(=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}\)

----------------------------

\(7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})\)

\(\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}\)

\(\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}\)

\(\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}\)

Cách 1:

\(E=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)

\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)

Cách 2:

\(E^2=(4+\sqrt{15})^2(\sqrt{10}-\sqrt{6})^2(4-\sqrt{15})=(4+\sqrt{15})(4-\sqrt{15})(4+\sqrt{15}).(16-4\sqrt{15})\)

\(=(16-15)(4+\sqrt{15})(4-\sqrt{15}).4=(16-15)(16-15).4=4\)

Vì $E>0$ nên $E=2$

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

= \(-33\sqrt{2}\)

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

= \(10-2\sqrt{21}+14\sqrt{21}\)

= \(10+12\sqrt{21}\)

a) Ta có: \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)

\(=\sqrt{3}\left(2+\sqrt{16}-\sqrt{25}-\sqrt{81}\right)\)

\(=\sqrt{3}\left(2+4-5-9\right)\)

\(=-8\sqrt{3}\)

b) Ta có: \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)

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

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

\(=7-5=2\)

c) Ta có: \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)

\(=\left(\sqrt{3}+1\right)\cdot\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)

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

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

\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)(Vì \(\sqrt{3}>1\))

\(=3-1=2\)

d) Ta có: \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)

\(=\sqrt{2}\cdot\left(5+\sqrt{9}-\sqrt{49}-\sqrt{144}\right)\)

\(=\sqrt{2}\cdot\left(5+3-7-12\right)\)

\(=-11\sqrt{2}\)

e) Ta có: \(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)

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

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

\(=3-5=-2\)

g) Ta có: \(\left(\sqrt{3}-1\right)\cdot\sqrt{4+2\sqrt{3}}\)

\(=\left(\sqrt{3}-1\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)

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

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

\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1>0\))

\(=3-1=2\)

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

a) Ta có: \(D=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\cdot\left(-\sqrt{2}\right)\)

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

\(=-2+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)

\(=-2+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=-2+\left|\sqrt{5}-1\right|\)

\(=-2+\sqrt{5}-1\)(Vì \(\sqrt{5}>1\))

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

b) Ta có: \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}\right)-\sqrt{75}\)

\(=2\sqrt{81}+4\sqrt{144}-5\sqrt{3}\)

\(=18+48-5\sqrt{3}\)

\(=66-5\sqrt{3}\)

c) Ta có: \(E=\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)

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

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

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

\(=\sqrt{2}\cdot\left(\sqrt{5}+\sqrt{3}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))

\(=\sqrt{2}\cdot\left(5-3\right)\)

\(=2\sqrt{2}\)

d) Ta có: \(P=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

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

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

\(=\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}}+\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}\)(Vì \(\sqrt{\frac{3}{2}}>\sqrt{\frac{1}{2}}>0\))

\(=2\sqrt{\frac{3}{2}}=\sqrt{4\cdot\frac{3}{2}}=\sqrt{6}\)

e) Ta có: \(M=-3\sqrt{50}+2\sqrt{98}-7\sqrt{72}\)

\(=\sqrt{2}\cdot\left(-3\cdot\sqrt{25}+2\cdot\sqrt{49}-7\cdot\sqrt{36}\right)\)

\(=\sqrt{2}\cdot\left(-15+14-42\right)\)

\(=-43\sqrt{2}\)

a) \(=\left|\sqrt{5}-3\right|+\sqrt{5-2\sqrt{20}+4}\)

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

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

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

\(=1\)

b)\(=\left(\sqrt{5}+1\right)\cdot\sqrt{2}\cdot\left(6-2\sqrt{5}\right)\cdot\sqrt{3+\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)

\(=16\)

c)\(=\frac{\sqrt{7}-5}{2}-\frac{6\left(\sqrt{7}+2\right)}{7-4}+\frac{\left(3-\sqrt{7}\right)}{9-7}+\frac{2\left(5-2\sqrt{7}\right)}{25-28}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6\left(\sqrt{7}+2\right)}{3}+\frac{3-\sqrt{7}}{2}-\frac{2\left(5-2\sqrt{7}\right)}{3}\)

\(=\frac{\sqrt{7}-\sqrt{7}-5+3}{2}-\frac{6\sqrt{7}+12+10-4\sqrt{7}}{3}\)

\(=-1-\frac{2\sqrt{7}+22}{3}\)

\(=\frac{-3-2\sqrt{7}-22}{3}\)

\(=-\frac{25+2\sqrt{7}}{3}\)