cmr\(\dfrac{\sqrt[4]{17+12\sqrt{2}}+\sqrt[4]{17-12\sqrt{2}}}{2}=\sqrt{2}\)
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AT
20 tháng 6 2021
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}}{\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2.\sqrt{2}.1+1^2}}{\sqrt{3^2+2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}+\dfrac{1}{\sqrt{2}+1}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2}+1-\sqrt{2}+1=2\)
11 tháng 8 2021
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 \(3\sqrt{12}>2\sqrt{16}\)
23 tháng 6 2017
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) = \(\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
= \(\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\) = \(\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\)
= \(\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-\left(3\sqrt{2}-4+3-2\sqrt{2}\right)}{9-8}\)
= \(\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{1}\)
= \(2\)
23 tháng 6 2018
\(1.\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}=\dfrac{\sqrt{3}+2+2-\sqrt{3}}{3-4}=-4\)\(2.\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=\dfrac{8+6\sqrt{2}+6\sqrt{2}-8}{16-18}=\dfrac{-12\sqrt{2}}{2}-6\sqrt{2}\)\(3.\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}=\sqrt{8-2.2\sqrt{2}.3+9}+\sqrt{8+2.2\sqrt{2}.3+9}=\sqrt{\left(2\sqrt{2}-3\right)^2}+\sqrt{\left(2\sqrt{2}+3\right)^2}=\text{|}2\sqrt{2}-3\text{|}+\text{|}2\sqrt{2}+3\text{|}=4\sqrt{2}\)
\(4.\sqrt{29-4\sqrt{7}}-\sqrt{29+4\sqrt{7}}=\sqrt{28-2.2\sqrt{7}.1+1}-\sqrt{28+2.2\sqrt{7}.1+1}=\sqrt{\left(2\sqrt{7}-1\right)^2}-\sqrt{\left(2\sqrt{7}+1\right)^2}=\text{|}2\sqrt{7}-1\text{|}-\text{|}2\sqrt{7}+1\text{|}=-2\)\(5.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7+2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{\text{|}\sqrt{7}+1\text{|}-\text{|}\sqrt{7}-1\text{|}}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2}\)
23 tháng 6 2018
1)
\(\dfrac{1}{\sqrt{3}-2}-\dfrac{1}{\sqrt{3}+2}\)
\(=\dfrac{\left(\sqrt{3}+2\right)-\left(\sqrt{3}-2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}\)
\(=\dfrac{4}{\left(\sqrt{3}\right)^2-2^2}\)
\(=\dfrac{4}{3-4}=-4\)
PT
3 tháng 7 2018
Phải là \(\sqrt{17+12\sqrt{2}}\) chớ bạn :<
\(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\)
\(=\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{1}\)
\(=2\)
18 tháng 9 2021
k: \(\sqrt[3]{\left(4-2\sqrt{3}\right)\left(\sqrt{3}-1\right)}\)
\(=\sqrt[3]{\left(\sqrt{3}-1\right)^3}\)
\(=\sqrt{3}-1\)
AT
7 tháng 7 2018
A: here
\(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}=\dfrac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}}-\dfrac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2\cdot3\cdot2\sqrt{2}+8}}=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}-\dfrac{\sqrt{2}+1}{\left(\sqrt{2}+1\right)^2}=\dfrac{1}{\sqrt{2}-1}-\dfrac{1}{\sqrt{2}+1}=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=\dfrac{2}{1}=2\)
7 tháng 7 2018
\(A=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{3+2\sqrt{3}+1}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{3-2\sqrt{3}+1}}=\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}=\dfrac{6\sqrt{2}-2\sqrt{6}+3\sqrt{6}-\sqrt{18}+6\sqrt{2}+2\sqrt{6}-3\sqrt{6}-\sqrt{18}}{9-3}=\dfrac{12\sqrt{2}-6\sqrt{2}}{6}=\sqrt{2}\) \(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}=\dfrac{\sqrt{2-2\sqrt{2}+1}}{\sqrt{9-2.3.2\sqrt{2}+8}}-\dfrac{\sqrt{2+2\sqrt{2}+1}}{\sqrt{9+2.3.2\sqrt{2}+8}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}=3\sqrt{2}+1-2\sqrt{2}-3\sqrt{2}+1+2\sqrt{2}=2\)
Ta có:
\(\dfrac{\sqrt[4]{17+12\sqrt{2}} +\sqrt[4]{17-12\sqrt{2}}}{2}\)
\(=\dfrac{\sqrt[4]{3^2+2.3.(2\sqrt{2})+\left(2\sqrt{2}\right)^2}+\sqrt[4]{3^2-2.3.(2\sqrt{2})+\left(2\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt[4]{\left(3+2\sqrt{2}\right)^2}+\sqrt[4]{\left(3-2\sqrt{2}\right)^2}}{2}\)
\(=\dfrac{\sqrt[4]{\left(2+2\sqrt{2}+1\right)^2}+\sqrt[4]{\left(2-2\sqrt{2}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt[4]{[\left(\sqrt{2}+1\right)^2]^2}+\sqrt[4]{[\left(\sqrt{2}-1\right)^2]^2}}{2}\)
\(=\dfrac{\sqrt[4]{\left(\sqrt{2}+1\right)^4}+\sqrt[4]{\left(\sqrt{2}-1\right)^4}}{2}\)
\(=\dfrac{\sqrt{2}+1+\sqrt{2}-1}{2}\)
\(=\dfrac{2\sqrt{2}}{2}\)
\(=\sqrt{2}\) (đpcm)