\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)

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

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

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

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) = \(\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)

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

= \(\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-\left(3\sqrt{2}-4+3-2\sqrt{2}\right)}{9-8}\)

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

= \(2\)

\(1.\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}=\dfrac{\sqrt{3}+2+2-\sqrt{3}}{3-4}=-4\)\(2.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{8+6\sqrt{2}+6\sqrt{2}-8}{16-18}=\dfrac{-12\sqrt{2}}{2}-6\sqrt{2}\)\(3.\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.3+9}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\text{|}2\sqrt{2}-3\text{|}+\text{|}2\sqrt{2}+3\text{|}=4\sqrt{2}\)
\(4.\sqrt{29-4\sqrt{7}}-\sqrt{29+4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}.1+1}-\sqrt{28+2.2\sqrt{7}.1+1}=\sqrt{\left(2\sqrt{7}-1\right)^2}-\sqrt{\left(2\sqrt{7}+1\right)^2}=\text{|}2\sqrt{7}-1\text{|}-\text{|}2\sqrt{7}+1\text{|}=-2\)\(5.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7+2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\text{|}\sqrt{7}+1\text{|}-\text{|}\sqrt{7}-1\text{|}}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}\)

1)

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

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

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

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

Phải là \(\sqrt{17+12\sqrt{2}}\) chớ bạn :<

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

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

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

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

\(=2\)

k: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)

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

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

A: here

\(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}=\dfrac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}}-\dfrac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2\cdot3\cdot2\sqrt{2}+8}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}-\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=\dfrac{2}{1}=2\)

\(A=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}=\dfrac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{9-3}=\dfrac{12\sqrt{2}-6\sqrt{2}}{6}=\sqrt{2}\) \(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}=\dfrac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2.3.2\sqrt{2}+8}}-\dfrac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2.3.2\sqrt{2}+8}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}=3\sqrt{2}+1-2\sqrt{2}-3\sqrt{2}+1+2\sqrt{2}=2\)