\(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)
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NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 8 2021
\(\sqrt{27-8\sqrt{11}}-\sqrt{11}=\sqrt{\left(4-\sqrt{11}\right)^2}-\sqrt{11}=\left|4-\sqrt{11}\right|-\sqrt{11}=4-2\sqrt{11}\)
\(\sqrt{48-16\sqrt{8}}-\sqrt{8}=\sqrt{\left(4\sqrt{2}-4\right)^2}-2\sqrt{2}=\left|4\sqrt{2}-4\right|-2\sqrt{2}=2\sqrt{2}-4\)
A
9 tháng 8 2021
\(\sqrt{27-8\sqrt{11}}-\sqrt{11}\\ =\sqrt{4^2-2.4.\sqrt{11}+\left(\sqrt{11}\right)^2}-\sqrt{11}\\ =\sqrt{\left(4-\sqrt{11}\right)^2}-\sqrt{11}\\ =\left|4-\sqrt{11}\right|-\sqrt{11}\\ =4-\sqrt{11}-\sqrt{11}=4-\left(-2\right)=6\)
\(\sqrt{48-16\sqrt{8}}-\sqrt{8}\\ =\sqrt{\left(4\sqrt{2}\right)^2-2.4\sqrt{2}.4+4^2}-\sqrt{8}\\ =\sqrt{\left(4\sqrt{2}-4\right)^2}-\sqrt{8}\\ =\left|4\sqrt{2}-4\right|-\sqrt{8}\\ =4\sqrt{2}-4-\sqrt{8}=4\sqrt{2}-4-3\sqrt{2}\\ =\sqrt{2}-4\)
24 tháng 6 2021
`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`
`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`
`=root{3}{4(1-sqrt3)(1+sqrt3)}`
`=root{3}{4(1-3)}=-2`
`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`
`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`
`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`
`=root{3}{9}`
24 tháng 6 2021
`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`
`=root{3}{(8sqrt5-16)(8sqrt5+16)}`
`=root{3}{320-256}`
`=root{3}{64}=4`
`b)root{3}{7-5sqrt2}-root{6}{8}`
`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`
`=root{3}{(1-sqrt2)^3}-sqrt2`
`=1-sqrt2-sqrt2=1-2sqrt2`
AD
2
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 9 2019
\(x=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)
\(\Rightarrow x^3=32+3\sqrt[3]{16^2-8^2.5}\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(\Rightarrow x^3=32-12x\)
\(\Rightarrow x^3+12x-32=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+16\right)=0\)
\(\Rightarrow x=2\)
Vậy \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
25 tháng 9 2019
x=16−853+16−853
⇒x3=32+3162−82.53(16−853+16+853)
⇒x3=32−12x
⇒x3+12x−32=0
⇒(x−2)(x2+2x+16)=0
⇒x=2
Vậy
NT
2
22 tháng 12 2015
\(\sqrt[3]{16-8\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+3\left(\sqrt{5}\right)^2-\left(\sqrt{5}\right)^3}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=\(1-\sqrt{5}\)
làm tương tự: \(\sqrt[3]{16+8\sqrt{5}}\)=\(1+\sqrt{5}\)
suy ra: a = 2
DM
0
HA
2
E
3 tháng 11 2018
\(x^3=16-8\sqrt{5}+16+8\sqrt{5}+3\sqrt[3]{\left(16-8\sqrt{5}\right)\left(16+8\sqrt{5}\right)}\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+5\sqrt{5}}\right)=32+3\sqrt[3]{256-320}.x=32-12x\)
<=> x3 +12x - 32 = 0
<=> x = 2
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: \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
\(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}=\sqrt[3]{\left(1+\sqrt{5}\right)^3}+\sqrt[3]{\left(1-\sqrt{5}\right)^3}=1+\sqrt{5}+1-\sqrt{5}=2\)
bn ơi làm thế nào để ra \(\left(1+\sqrt{5}\right)^3\)