a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

\(a,\sqrt{\sqrt{17+12\sqrt{2}}}\)

\(=\sqrt{\sqrt{8+12\sqrt{2}+9}}\)

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

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

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

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

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

\(b,\sqrt{4+2\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{3+2\sqrt{3}+1}-\sqrt{12-12\sqrt{3}+9}\)

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

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

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

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

chúc bn học tốt

Bài 1:

\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{8-2\sqrt{8.9}+9}=\sqrt{(\sqrt{8}-\sqrt{9})^2}\)

\(=|\sqrt{8}-\sqrt{9}|=3-2\sqrt{2}\)

\(\Rightarrow a=3; b=-\sqrt{2}\)

\(\Rightarrow a^2+b^2=9+2=11\)

Bài 1: 

Ta có: \(\sqrt{17-12\sqrt{2}}=a+b\sqrt{2}\)

\(\Leftrightarrow a+b\sqrt{2}=3-2\sqrt{2}\)

Suy ra: a=3; b=-2

\(\Leftrightarrow a^2+b^2=3^2+\left(-2\right)^2=9+4=13\)

Bài 2: 

a) Ta có: \(\dfrac{\sqrt{a}-1}{a\sqrt{a}+\sqrt{a}-a}:\dfrac{1}{a^2+a}\)

\(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{a\left(a+1\right)}{1}\)

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

\(\sqrt{6-4\sqrt{2}}\)\(+\sqrt{22-12\sqrt{2}}\)

\(=\sqrt{4-4\sqrt{2}+2}\)\(+\sqrt{18-12\sqrt{2}+4}\)

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

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

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

\(=0+2\sqrt{2}\)\(=2\sqrt{2}\)

\(\sqrt{17-12\sqrt{2}}\)\(+\sqrt{9+4\sqrt{2}}\)

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

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

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

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

\(=4+0=4\)

\(\sqrt{17+12\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{3^2+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

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

\(=4\sqrt{2}\)

\(\sqrt{17+12\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}-\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\)

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

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

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

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

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

\(=-\sqrt{2}\)

a: \(=\sqrt{8+2\cdot2\sqrt{2}\cdot\sqrt{5}+5}+\sqrt{8-2\cdot2\sqrt{2}\cdot\sqrt{5}+5}\)

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

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

b: \(=2\cdot\sqrt{17-3\sqrt{32}}\)

\(=2\cdot\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}\)

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