Lời giải:

a)

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}\)

\(=2\sqrt{5}-\sqrt{25}.\sqrt{5}-\sqrt{16}.\sqrt{5}+\sqrt{121}.\sqrt{5}\)

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

b)

\(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}=\frac{\sqrt{20}(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}+\frac{8(1+\sqrt{5})}{(1-\sqrt{5})(1+\sqrt{5})}\)

\(=\sqrt{20}+\frac{8(1+\sqrt{5})}{1-5}=2\sqrt{5}-2(1+\sqrt{5})=-2\)

9.

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\)

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

\(=-\sqrt{5}\)

10.

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\)

\(=5\sqrt{3}-\sqrt{5+\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2+\dfrac{2}{3}}+6\sqrt{3}\)

\(=11\sqrt{3}-\sqrt{\dfrac{16}{3}}+\dfrac{9}{2}\sqrt{\dfrac{8}{3}}\)

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

\(=\dfrac{29\sqrt{3}}{3}+3\sqrt{6}\)

\(\sqrt{20}+2\sqrt{45}+\sqrt{125}-3\sqrt{80}\\ =2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}=\sqrt{5}\)

\(\sqrt{75}-\sqrt{5\dfrac{1}{3}}+\dfrac{9}{2}\sqrt{2\dfrac{2}{3}}+2\sqrt{27}\\ =5\sqrt{3}-\dfrac{4\sqrt{3}}{3}+3\sqrt{6}+6\sqrt{3}\\ =\dfrac{15\sqrt{3}-4\sqrt{3}+6\sqrt{6}+18\sqrt{3}}{3}\\ =\dfrac{29\sqrt{3}+6\sqrt{6}}{3}\)

Lời giải:

\(M=\frac{2\sqrt{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+\sqrt{(\sqrt{20}+1)^2}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{(\sqrt{5}+1)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-(\sqrt{5}+1)}}{\sqrt{2}(\sqrt{5}-1)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)

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

Lời giải:

\(M=\frac{2\sqrt{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+\sqrt{(\sqrt{20}+1)^2}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)

\(=\frac{2\sqrt{4-\sqrt{(\sqrt{5}+1)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{2\sqrt{4-(\sqrt{5}+1)}}{\sqrt{2}(\sqrt{5}-1)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)

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

\(A=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{20+4\sqrt{5}+1}}}}{\sqrt{10}-\sqrt{2}}\)\(=\dfrac{2\sqrt{4-\sqrt{5+\sqrt{\left(2\sqrt{5}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5+\left(2\sqrt{5}+1\right)}}}{\sqrt{10}-\sqrt{2}}=\)\(=\dfrac{2\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{4-\sqrt{5}-1}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{6-2\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{5-2\sqrt{5}+1}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{2}.\left(\sqrt{5}-1\right)}{\sqrt{10}-\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{10}-\sqrt{2}}=1\)

\(A=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\\ =\dfrac{4\sqrt{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\dfrac{4\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}\\ =\dfrac{2\sqrt{\left(\sqrt{10}-\sqrt{2}\right)^2}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)}{\sqrt{10}-\sqrt{2}}\\ =2\)

GIẢI 

\(M=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{20+1+2\sqrt{20.1}}}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{\left(\sqrt{20}+1\right)^2}}}}{\sqrt{10}-\sqrt{2}}\)

        \(=\frac{\sqrt[2]{4-\sqrt{5+\sqrt{20}+1}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\sqrt{5+1+2\sqrt{5}}}}{\sqrt{10}-\sqrt{2}}\)

         \(=\frac{\sqrt[2]{4-\sqrt{\left(\sqrt{5}+1\right)^2}}}{\sqrt{10}-\sqrt{2}}=\frac{\sqrt[2]{4-\left(\sqrt{5}+1\right)}}{\sqrt{2}\left(\sqrt{5}-1\right)}=\frac{\sqrt{2}.\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}\)

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

Chúc bạn học tốt !!!

1 bài muốn chết tới 3 bài tớ vào quan tài nằm trước

1,

\(2\sqrt{5}-\sqrt{125}-\sqrt{80}\\ =2\sqrt{5}-\sqrt{25\cdot5}-\sqrt{16\cdot5}\\ =2\sqrt{5}-5\sqrt{5}-4\sqrt{5}\\ =-7\sqrt{5}\)

2,

\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}\\ =3\sqrt{2}-\sqrt{4\cdot2}+\sqrt{25\cdot2}-4\sqrt{16\cdot2}\\ =3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}\\=-10\sqrt{2}\)

3,

\(\sqrt{18}-3\sqrt{80}-2\sqrt{50}+2\sqrt{45}\\ =\sqrt{9\cdot2}-3\sqrt{16\cdot5}-2\sqrt{25\cdot2}+2\sqrt{9\cdot5}\\ =3\sqrt{2}-12\sqrt{5}-10\sqrt{2}+6\sqrt{5}\\ =-7\sqrt{2}-6\sqrt{5}\)

4,

\(\sqrt{27}-2\sqrt{3}+2\sqrt{48}-3\sqrt{75}\\ =\sqrt{9\cdot3}-2\sqrt{3}+2\sqrt{16\cdot3}-3\sqrt{25\cdot2}\\ =3\sqrt{3}-2\sqrt{3}+8\sqrt{3}-15\sqrt{3}\\ =-6\sqrt{3}\)

5,

\(3\sqrt{2}-4\sqrt{18}+\sqrt{32}-\sqrt{50}\\ =3\sqrt{2}-4\sqrt{9\cdot2}+\sqrt{16\cdot2}-\sqrt{25\cdot2}\\ =3\sqrt{2}-12\sqrt{2}+4\sqrt{2}-5\sqrt{2}\\ =-10\sqrt{2}\)

6,

\(2\sqrt{3}-\sqrt{75}+2\sqrt{12}-\sqrt{147}\\ =2\sqrt{3}-\sqrt{25\cdot3}+2\sqrt{4\cdot3}-\sqrt{49\cdot3}\\ =2\sqrt{3}-5\sqrt{3}+4\sqrt{3}-7\sqrt{3}\\ =-6\sqrt{3}\)

7,

\(\sqrt{20}-2\sqrt{45}-3\sqrt{80}+\sqrt{125}\\ =\sqrt{4\cdot5}-2\sqrt{9\cdot5}-3\sqrt{16\cdot5}+\sqrt{25\cdot5}\\ =2\sqrt{5}-6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\\ =-11\sqrt{5}\)

8,

\(6\sqrt{12}-\sqrt{20}-2\sqrt{27}+\sqrt{125}\\ =6\sqrt{4\cdot3}-\sqrt{4\cdot5}-2\sqrt{9\cdot3}+\sqrt{25\cdot5}\\ =12\sqrt{3}-2\sqrt{5}-6\sqrt{3}+5\sqrt{5}\\ =6\sqrt{3}+3\sqrt{5}\\ =3\left(2\sqrt{3}+\sqrt{5}\right)\)

9,

\(4\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\\ =4\sqrt{4\cdot6}-2\sqrt{9\cdot6}+3\sqrt{6}-\sqrt{25\cdot6}\\ =8\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}=0\)

10,

\(2\sqrt{18}-3\sqrt{80}-5\sqrt{147}+5\sqrt{245}-3\sqrt{98}\\ =2\sqrt{9\cdot2}-3\sqrt{16\cdot5}-5\sqrt{49\cdot3}+5\sqrt{49\cdot5}-3\sqrt{49\cdot2}\\ =6\sqrt{2}-12\sqrt{5}-35\sqrt{3}+35\sqrt{5}-21\sqrt{2}\\ =-15\sqrt{2}-35\sqrt{3}+23\sqrt{5}\)

a. \(\frac{26}{5-2\sqrt{3}}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{\left(5-2\sqrt{3}\right)\left(5+2\sqrt{3}\right)}\)=\(\frac{26\cdot\left(5+2\sqrt{3}\right)}{5^2-\left(2\sqrt{3}\right)^2}=\frac{26\cdot\left(5+2\sqrt{3}\right)}{13}=2\cdot\left(5+2\sqrt{3}\right)=10+4\sqrt{3}\)

b.\(\frac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}=\frac{\sqrt{3}\cdot\left(3\sqrt{3}-2\right)}{\sqrt{2}\cdot\left(3\sqrt{3}-2\right)}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}\)

c.\(\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\sqrt{5}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}\cdot\left(2\sqrt{2}-\sqrt{5}\right)}=\frac{\sqrt{5}}{\sqrt{2}}=\frac{\sqrt{10}}{2}\)

d.\(2\sqrt{5}-\sqrt{125}-\sqrt{80}+\sqrt{605}=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}\)=\(4\sqrt{5}\)