\(\sqrt[5]{32=?}\)
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N
25 tháng 10 2021
\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 8 2021
\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}=6\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+2\sqrt{2}\right)^2}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}=2\sqrt{5}+4\sqrt{2}\)
21 tháng 8 2021
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
b: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c: Ta có: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
d: Ta có: \(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{13-4\sqrt{10}}+\sqrt{53+4\sqrt{90}}\)
\(=2\sqrt{2}-\sqrt{5}+3\sqrt{5}+2\sqrt{2}\)
\(=2\sqrt{5}+4\sqrt{2}\)
30 tháng 7 2021
a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)
\(=\sqrt{5}-1\)
b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)
\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)
\(=2\sqrt{2}\)
30 tháng 10 2023
a: \(\dfrac{\sqrt{50}-\sqrt{32}+\sqrt{8}}{\sqrt{2}}\)
\(=\dfrac{5\sqrt{2}-4\sqrt{2}+2\sqrt{2}}{\sqrt{2}}\)
\(=\dfrac{3\sqrt{2}}{\sqrt{2}}=3\)
b: \(\dfrac{4}{\sqrt{5}-1}-5\sqrt{\dfrac{1}{5}}\)
\(=\dfrac{4\left(\sqrt{5}+1\right)}{5-1}-\sqrt{5}\)
\(=\sqrt{5}+1-\sqrt{5}\)
=1
M
14 tháng 8 2020
a, \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
= \(\sqrt{3^2-2.3.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{6^2-2.6.\sqrt{6}+\left(\sqrt{6}\right)^2}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(6-\sqrt{6}\right)^2}\)
= \(\left|3-\sqrt{6}\right|+\left|6-\sqrt{6}\right|\)
= \(3-\sqrt{6}+6-\sqrt{6}\)
= \(9-2\sqrt{6}\)
b. Đặt B = \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
Nhận xét : B > 0 , bình phương hai vế ta được :
\(B^2=\left(\sqrt{17-3\sqrt{32}}\right)^2+\left(\sqrt{17+3\sqrt{32}}\right)^2\)
\(B^2=17-3\sqrt{32}+17+3\sqrt{32}+2\sqrt{\left(17-3\sqrt{32}\right)\left(17+3\sqrt{32}\right)}\)
\(B^2=34+2\sqrt{17^2-\left(3\sqrt{32}\right)^2}\)
\(B^2=34+2\sqrt{289-288}\)
\(B^2=34+2=36\)
=> \(B=\pm\sqrt{36}\) mà B > 0 nên \(B=\sqrt{36}=6\)
c, Đặt C = \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
Nhận xét : C > 0 , bình phương hai vế ta đươc :
\(C^2=\left(\sqrt{49-5\sqrt{96}}\right)^2+\left(\sqrt{49+5\sqrt{96}}\right)^2\)
\(C^2=49-5\sqrt{96}+49+5\sqrt{96}+2\sqrt{\left(49-5\sqrt{96}\right)\left(49+5\sqrt{96}\right)}\)
\(C^2=98+2\sqrt{49^2-\left(5\sqrt{96}\right)^2}\)
\(C^2=98+2\sqrt{2401-2400}\)
\(C^2=98+2=100\)
=> \(C=\pm\sqrt{100}\) mà C > 0 nên \(C=\sqrt{100}=10\)
14 tháng 8 2020
a) Ta có: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot3\sqrt{3}\cdot2\sqrt{2}+8}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|3\sqrt{3}-2\sqrt{2}\right|\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)(Vì \(\left\{{}\begin{matrix}3>\sqrt{6}\\3\sqrt{3}>2\sqrt{2}\end{matrix}\right.\))
b) Ta có: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\frac{\sqrt{34-6\sqrt{32}}+\sqrt{34+6\sqrt{32}}}{\sqrt{2}}\)
\(=\frac{\sqrt{18-2\cdot3\sqrt{2}\cdot4+16}+\sqrt{18+2\cdot3\sqrt{2}\cdot4+16}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{2}-4\right)^2}+\sqrt{\left(3\sqrt{2}+4\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|3\sqrt{2}-4\right|+\left|3\sqrt{2}+4\right|}{\sqrt{2}}\)
\(=\frac{3\sqrt{2}-4+3\sqrt{2}+4}{\sqrt{2}}\)(Vì \(3\sqrt{2}>4>0\))
\(=\frac{6\sqrt{2}}{\sqrt{2}}=6\)
H9
HT.Phong (9A5)
CTVHS
17 tháng 7 2023
1) \(\sqrt{6+4\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{3^2-2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|2+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\)
\(=2+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}-1\)
2) \(\sqrt{21-4\sqrt{5}}+\sqrt{21+4\sqrt{5}}\)
\(=\sqrt{20-4\sqrt{5}+1}+\sqrt{20+4\sqrt{5}+1}\)
\(=\sqrt{\left(2\sqrt{5}\right)^2-2\sqrt{5}\cdot2\cdot1+1^2}+\sqrt{\left(2\sqrt{5}\right)^2+2\sqrt{5}\cdot2\cdot1-1^2}\)
\(=\sqrt{\left(2\sqrt{5}-1\right)^2}+\sqrt{\left(2\sqrt{5}+1\right)^2}\)
\(=\left|2\sqrt{5}-1\right|+\left|2\sqrt{5}+1\right|\)
\(=2\sqrt{5}-1+2\sqrt{5}+1\)
\(=4\sqrt{5}\)
AP
19 tháng 10 2018
a, \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)
b,\(\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{25-5}=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{20}=\dfrac{60}{20}=3\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 8 2023
\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{17-12\sqrt{2}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot\sqrt{7}+\left(\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{\sqrt{\left(2\sqrt{2}-3\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{\left(3-2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}+\sqrt{7}\right)^2}}\)
\(=\dfrac{2\sqrt{2}+3}{2\sqrt{2}-3}+\dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}-\dfrac{1}{\sqrt{5}+\sqrt{7}}\)
\(=\dfrac{\left(2\sqrt{2}+3\right)^2}{\left(2\sqrt{2}+3\right)\left(2\sqrt{2}-3\right)}+1-\dfrac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)
\(=17-12\sqrt{2}+1-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{2\cdot\left(18-12\sqrt{2}\right)}{2}-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)
\(=\dfrac{36-24\sqrt{2}-\sqrt{5}+\sqrt{7}}{2}\)
I
1
16 tháng 8 2021
\(\sqrt[3]{53\sqrt{5}+124}+\sqrt[3]{32\sqrt{5}-72}\)
\(=\sqrt[3]{\left(\sqrt{5}\right)^3+3.5.4+3.\sqrt{5}.4+4^3}+\sqrt[3]{\left(\sqrt{5}\right)^3-3.5.3+3.\sqrt{5}.3^2-3^3}\)
\(=\sqrt[3]{\left(\sqrt{5}+4\right)^3}+\sqrt[3]{\left(\sqrt{5}-3\right)^3}\)
\(=\sqrt{5}+4+\sqrt{5}-3\)
\(=2\sqrt{5}+1\)
\(\sqrt[5]{32}\)= 2
\(\sqrt[5]{32}\)\(=2\)
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