\(\sqrt{\frac{\text{}\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}\). ( \(3\sqrt{2}\)+ \(\sqrt{14}\))
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30 tháng 10 2020
a) Ta có: \(\frac{7\sqrt{2}+2\sqrt{7}}{\sqrt{14}}-\frac{5}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\sqrt{14}\left(\sqrt{7}+\sqrt{2}\right)}{\sqrt{14}}-\frac{5\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{2}\right)-5\left(\sqrt{7}-\sqrt{5}\right)}{2}\)
\(=\frac{2\sqrt{7}+2\sqrt{2}-5\sqrt{7}+5\sqrt{5}}{2}\)
\(=\frac{2\sqrt{2}-3\sqrt{7}+5\sqrt{5}}{2}\)
b) Ta có: \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{\sqrt{2}\left(6+2\sqrt{5}\right)}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(6-2\sqrt{5}\right)}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\left|\sqrt{5}+1\right|}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left|\sqrt{5}-1\right|}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\left(\sqrt{5}+1\right)}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left(\sqrt{5}-1\right)}\)(Vì \(\sqrt{5}>1>0\))
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{10}+\sqrt{2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{5\sqrt{2}+\sqrt{10}}+\frac{6\sqrt{2}-2\sqrt{10}}{5\sqrt{2}-\sqrt{10}}\)
\(=\frac{6+2\sqrt{5}}{5+\sqrt{5}}+\frac{6-2\sqrt{5}}{5-\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\sqrt{5}\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{5}}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}}=2\)
c) Đặt \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
Ta có: \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
\(\Leftrightarrow A^3=32-12\cdot\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(=32-12A\)
\(\Leftrightarrow A^3+12A-32=0\)
\(\Leftrightarrow A^3-2A^2+2A^2-4A+16A-32=0\)
\(\Leftrightarrow A^2\left(A-2\right)+2A\left(A-2\right)+16\left(A-2\right)=0\)
\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+16\right)=0\)
mà \(A^2+2A+16>0\)
nên A-2=0
hay A=2
Vậy: \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
LT
0
D
1
30 tháng 11 2016
A=\(\sqrt{\frac{\sqrt{5}}{8\sqrt{5}+3\sqrt{35}}}.3\sqrt{2}+\sqrt{14}=\sqrt{\frac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}.\sqrt{2}\left(3+\sqrt{7}\right)\)
\(8>3.\sqrt{7}\Rightarrow8-3\sqrt{7}>0\left(lienhop\right)\left(8-3\sqrt{7}\right)\)
\(A=\sqrt{\left(8-3.\sqrt{7}\right)}.\sqrt{2}\left(3+\sqrt{7}\right)\)
\(A=\sqrt{\left(16-2.3.\sqrt{7}\right)}.\left(3+\sqrt{7}\right)\)
\(A=\sqrt{3^2-2.3.\sqrt{7}+\left(\sqrt{7}\right)^2}.\left(3+\sqrt{7}\right)\)
\(A=\sqrt{\left(3-\sqrt{7}\right)^2}\left(3+\sqrt{7}\right)\)
\(3-\sqrt{7}>0\)
\(\Rightarrow A=\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)=9-7=2\)
CC
0
17 tháng 12 2015
\(A=\sqrt{\frac{1}{8+3\sqrt{7}}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(A=\sqrt{\frac{2}{16+6\sqrt{7}}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(A=\frac{\sqrt{2}}{3+\sqrt{7}}\left(3+\sqrt{7}\right)\sqrt{2}\)
\(A=2\)
D
1
19 tháng 10 2021
\(=\sqrt{\dfrac{\sqrt{5}}{\sqrt{5}\left(8+3\sqrt{7}\right)}}\cdot\sqrt{2}\left(3+\sqrt{7}\right)\\ =\sqrt{\dfrac{2\left(3+\sqrt{7}\right)^2}{8+3\sqrt{7}}}=\sqrt{\dfrac{32+12\sqrt{7}}{8+3\sqrt{7}}}\\ =\sqrt{\dfrac{4\left(8+3\sqrt{7}\right)}{8+3\sqrt{7}}}=\sqrt{4}=2\)
5 tháng 10 2021
b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)
\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)
\(=4\left(7+3\sqrt{5}\right)\)
\(=28+12\sqrt{5}\)
AH
Akai Haruma
Giáo viên
5 tháng 10 2021
Lời giải:
a.
$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$
$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$
$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$
$=2\sqrt{5}-5\sqrt{10}$
$\Rightarrow A=\sqrt{10}-5\sqrt{5}$
b.
$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$
$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$
$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$
$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$
$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$
$\Rightarrow B=28+12\sqrt{5}$
c.
$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}
$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$
$=(7-5)(6-\sqrt{35})$
$=2(6-\sqrt{35})=12-2\sqrt{35}$
\(=\sqrt{\frac{\sqrt{5}\left(8\sqrt{5}-3\sqrt{35}\right)}{\left(8\sqrt{5}+3\sqrt{35}\right)\left(8\sqrt{5}-3\sqrt{35}\right)}}\)\(\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{\frac{40-15\sqrt{7}}{5}}.\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\sqrt{8-3\sqrt{7}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\sqrt{2}\sqrt{8-3\sqrt{7}}}{\sqrt{2}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\sqrt{16-3\sqrt{7}}}{\sqrt{2}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\sqrt{\left(3-\sqrt{7}\right)^2}}{\sqrt{2}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(=\frac{\left(3-\sqrt{7}\right)}{\sqrt{2}}.\sqrt{2}\left(3+\sqrt{7}\right)\)
\(=9-7\)
\(=2\)