`a)\sqrt{9-4sqrt5}-sqrt5`

`=sqrt{5-2.2sqrt5+4}-sqrt5`

`=sqrt{(sqrt5-2)^2}-sqrt5`

`=|\sqrt5-2|-sqrt5`

`=sqrt5-2-sqrt5=-2`

`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`

`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`

`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`

`=|2-sqrt3|+|sqrt3-1|`

`=2-sqrt3+sqrt3-1=1`

`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`

`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`

`=sqrtx+7`

`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`

`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`

`=sqrt3+1-2sqrt3-1=-sqrt3`

`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)

Lời giải:

a)
\((\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}})^2=5-2\sqrt{5}+5+2\sqrt{5}+2\sqrt{(5-2\sqrt{5})(5+2\sqrt{5})}\)

\(=10+2\sqrt{5^2-(2\sqrt{5})^2}=10+2\sqrt{5}\)

\(\Rightarrow \sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}=\sqrt{10+2\sqrt{5}}\)

b)

\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)

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

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

c)

\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)

\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)

\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)

Lời giải:

a)
\((\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}})^2=5-2\sqrt{5}+5+2\sqrt{5}+2\sqrt{(5-2\sqrt{5})(5+2\sqrt{5})}\)

\(=10+2\sqrt{5^2-(2\sqrt{5})^2}=10+2\sqrt{5}\)

\(\Rightarrow \sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}=\sqrt{10+2\sqrt{5}}\)

b)

\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)

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

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

c)

\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)

\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)

\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)

a: \(=2\sqrt{2}+1-3=2\sqrt{2}-2\)

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

c: \(=2-\sqrt{3}+\sqrt{3}-1=1\)

\(a,=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}\)

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

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

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

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

\(b,=\sqrt{6-2\sqrt{3+\sqrt{12+2\sqrt{12}+1}}}\)

\(=\sqrt{6-2\sqrt{3+\sqrt{12}+1}}\)

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

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

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

\(c,=\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{4+2.2\sqrt{3}+3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{25-2.5\sqrt{3}+3}}\)

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

\(d,=\sqrt{23-6\sqrt{10+4\sqrt{2-2\sqrt{2}+1}}}\)

\(=\sqrt{23-6\sqrt{6+4\sqrt{2}}}\)

\(=\sqrt{23-6\sqrt{4+2.2\sqrt{2}+2}}\)

\(=\sqrt{23-6\sqrt{\left(2+\sqrt{2}\right)^2}}\)

\(=\sqrt{23-12-6\sqrt{2}}=\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{9-2.3\sqrt{2}+2}=3-\sqrt{2}\)

a) Ta có: \(\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

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

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

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

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

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

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

c) Ta có: \(\sqrt{\sqrt{3}+\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{\sqrt{3}+\sqrt{48-10\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{\sqrt{3}+\sqrt{28-10\sqrt{3}}}\)

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

\(=\sqrt{5}\)

d) Ta có: \(\sqrt{23-6\sqrt{10+4\sqrt{3-2\sqrt{2}}}}\)

\(=\sqrt{23-6\sqrt{10+4\left(\sqrt{2}-1\right)}}\)

\(=\sqrt{23-6\sqrt{6-4\sqrt{2}}}\)

\(=\sqrt{23-6\left(2-\sqrt{2}\right)}\)

\(=\sqrt{11+6\sqrt{2}}\)

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

g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)

h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)

l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)

j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)

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

\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)

\(=\sqrt{12}+1=2\sqrt{3}+1\)

\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)

\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)

\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)

2) biến đổi khúc sau như câu 1:

\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)

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

4) Ta có: \(\sqrt{30-2\sqrt{16+6\sqrt{11+4\sqrt{4-2\sqrt{3}}}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{11+4\left(\sqrt{3}-1\right)}}}\)

\(=\sqrt{30-2\sqrt{16+6\sqrt{7+4\sqrt{3}}}}\)

\(=\sqrt{30-2\sqrt{16+6\left(2+\sqrt{3}\right)}}\)

\(=\sqrt{30-2\sqrt{28+6\sqrt{3}}}\)

\(=\sqrt{30-2\left(3\sqrt{3}+1\right)}\)

\(=\sqrt{28-6\sqrt{3}}=3\sqrt{3}-1\)

5) Ta có: \(\dfrac{\left(5\sqrt{3}+\sqrt{50}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{5\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{\sqrt{75}-5\sqrt{2}}\)

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

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...

Lời giải:

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

b. \(\sqrt{7-4\sqrt{3}}=\sqrt{4-2\sqrt{4}.\sqrt{3}+3}=\sqrt{(\sqrt{4}-\sqrt{3})^2}=\sqrt{4}-\sqrt{3}=2-\sqrt{3}\)

c.

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

\(=\sqrt{(\sqrt{2}-1)^2}-\sqrt{(\sqrt{4}-\sqrt{2})^2}\)

\(=|\sqrt{2}-1|-|\sqrt{4}-\sqrt{2}|=\sqrt{2}-1-(2-\sqrt{2})=2\sqrt{2}-3\)

d.

\(=\sqrt{13+30\sqrt{2+\sqrt{(\sqrt{8}+1)^2}}}=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}=\sqrt{13+30\sqrt{(\sqrt{2}+1)^2}}\)

\(=\sqrt{13+30(\sqrt{2}+1)}=\sqrt{43+30\sqrt{2}}=\sqrt{18+2\sqrt{18.25}+25}\)

\(=\sqrt{(\sqrt{18}+\sqrt{25})^2}=\sqrt{18}+\sqrt{25}=5+3\sqrt{2}\)

a) \(\sqrt{6-2\sqrt{5}}=\sqrt{5}-1\)

b) \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

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

d) Ta có: \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+1+2\sqrt{2}}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

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

\(1.\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\text{|}\sqrt{5}-\sqrt{2}\text{|}-\text{|}\sqrt{5}+\sqrt{2}\text{|}=-2\sqrt{2}\)\(2.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{|}2\sqrt{2}+\sqrt{5}\text{|}+\text{|}2\sqrt{2}-\sqrt{5}\text{|}=4\sqrt{2}\)\(3.\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{5}\right)\text{|}\sqrt{5}-\sqrt{2}\text{|}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)