a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)

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

=7-2

=5

d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)

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

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

\(=4\sqrt{7}\)

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.

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

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

a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

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

\(=10\sqrt{5}\)

b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

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

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

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

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

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

=7-5=2

d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

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

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

a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)

\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)

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

\(=10\sqrt{5}\)

b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)

\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)

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

\(=10\)

c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

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

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

\(=7-5=2\)

d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)

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

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

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

\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))

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

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

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: \(=2\sqrt{5}-2\sqrt{5}+9\sqrt{5}-30\sqrt{5}=-21\sqrt{5}\)

b: \(=2\sqrt{7}-6\sqrt{7}-\dfrac{3}{4}\sqrt{7}-8\sqrt{7}=-\dfrac{51}{4}\sqrt{7}\)

a)                  \(A=\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}\)

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

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

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

\(\Rightarrow\)\(A=\frac{\sqrt{5}-1}{\sqrt{2}}\)

b) tương tự câu a

c) \(\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}\)

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

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

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

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

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

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

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

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

a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)

b) \(\sqrt{16}+\sqrt{64}=4+8=12\)

c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)

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

\(a.\sqrt{4-\sqrt{15}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{5-2.\sqrt{5}.\sqrt{3}+3}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{3}-1}{\sqrt{2}}=\dfrac{\sqrt{5}-2\sqrt{3}-1}{\sqrt{2}}\)

\(b.\sqrt{4+\sqrt{15}}+\sqrt{7-\sqrt{45}}=\dfrac{\sqrt{5+2\sqrt{5}.\sqrt{3}+3}+\sqrt{9-2.3\sqrt{5}+5}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}+3-\sqrt{5}}{\sqrt{2}}=\dfrac{3+\sqrt{3}}{\sqrt{2}}\)

\(c.\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}-\sqrt{6-2\sqrt{5+\sqrt{13-4\sqrt{3}}}}=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}-\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}=\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)