9*9+2+80=
K
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GH
21 tháng 7 2023
Nháp:
\(9\pm\sqrt{80}=9\pm4\sqrt{5}=\dfrac{72\pm32\sqrt{5}}{8}=\left(\dfrac{3\pm\sqrt{5}}{2}\right)^3\)
\(\Rightarrow \sqrt[3]{9+\sqrt{80}}=\dfrac{3+\sqrt{5}}{2}\); \(\Rightarrow \sqrt[3]{9-\sqrt{80}}=\dfrac{3-\sqrt{5}}{2}\)
\(S=\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\\ S=2+\sqrt{3}+\dfrac{3+\sqrt{5}}{2}+\dfrac{3-\sqrt{5}}{2}\\ S=2+\sqrt{3}+3\\ S=5+\sqrt{3}\)
23 tháng 9 2021
\(x=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}=\dfrac{1}{\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}}\)
Đặt \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)\(\Leftrightarrow A^3=18+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow A^3=18+3A\sqrt[3]{1}\\ \Leftrightarrow A^3-3A-18=0\\ \Leftrightarrow A=3\\ \Leftrightarrow X=\dfrac{1}{3}\\ \Leftrightarrow Q=\left[3\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^2-1\right]^{2021}=\left(\dfrac{1}{9}-\dfrac{1}{9}-1\right)^{2021}=\left(-1\right)^{2021}=-1\)
25 tháng 8 2017
Đặt A = \(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)=> \(A^3=18+3A\Leftrightarrow A^3-3A-18=0\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\Leftrightarrow A-3=0\Leftrightarrow A=3\)
\(\dfrac{\sqrt[3]{26+15\sqrt{3}}\left(2-\sqrt{3}\right)}{\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}}=\dfrac{\sqrt[3]{\left(2+\sqrt{3}\right)^3}\left(2-\sqrt{3}\right)}{3}=\dfrac{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}{3}=\dfrac{1}{3}\)
A
2
PT
2
24 tháng 4 2015
a) (1,1x-9).2/5=80%=4/5
1,1x-9=4/5:2/5=4/5*5/2=2
1,1x=2+9
1,1x=11
x=11/1,1
x=10
24 tháng 4 2015
(1,1x-9).2/5=80%=4/5
1,1x-9=4/5:2/5=4/5*5/2=2
1,1x=2+9
1,1x=11
x=11/1,1
x=10
T
11 tháng 7 2017
Ta có \(\sqrt[3]{26+15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}\)
\(=\sqrt[3]{2^3+3.2^2\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}\)
\(=2+\sqrt{3}\)
Đặt \(x=\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\)
Ta có \(x^3=\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)^3\)
\(=9+\sqrt{80}+9-\sqrt{80}+3.\left(\sqrt[3]{9+\sqrt{80}}\right)^2\left(\sqrt[3]{9-\sqrt{80}}\right)+3.\left(\sqrt[3]{9-\sqrt{80}}\right)^2\left(\sqrt[3]{9+\sqrt{80}}\right)\)
\(=18+3\sqrt[3]{9+\sqrt{80}}.\sqrt[3]{9-\sqrt{80}}\left(\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}\right)\)
\(=18+3\sqrt[3]{9^2-80}.x\)
\(=18+3x\)
Vậy \(x^3=18+3x\)
\(\Leftrightarrow x^3-3x-18=0\)
Vậy x = 3
Do đó \(M=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)+3=2^2-3+3=4\)
Vậy M = 4.