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a: \(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot A\)
=>A^3-3A-18=0
=>A=3
b: \(B=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
=>\(B^3=5\sqrt{2}+7-5\sqrt{2}+7+3B\)
=>B^3-3B-14=0
=>B=2,82
c: \(C^3=20+14\sqrt{2}-14\sqrt{2}+20-6C\)
=>C^3+6C-40=0
=>C=2,84
Đặt \(x=\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\)
\(\Rightarrow x^3=40+3\sqrt[3]{\left(20+14\sqrt[]{2}\right)\left(20-14\sqrt[]{2}\right)}.\left(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}\right)\)
\(\Rightarrow x^3=40+6x\)
\(\Rightarrow x^3-6x-40=0\)
\(\Rightarrow\left(x-4\right)\left(x^2+4x+10\right)=0\)
\(\Rightarrow x=4\)
Vậy \(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}=4\)
A = \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
=> A3 = 40 + 6A
<=> A = 4
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
`x=root{3}{14sqrt2+20}+sqrt{-14sqrt2+20}`
`<=>x^3=14sqrt2+20-14sqrt2+20+3root{3}{(14sqrt2+20)(20-14sqrt2)}(root{3}{14sqrt2+20}+sqrt{-14sqrt2+20})`
`<=>x^3=40+3root{3}{400-392}.x`
`<=>x^3=40+6x`
`<=>x^3-6x=40`
\(x=\sqrt[3]{30+14\sqrt{2}}-\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left[2^3+3.2^2.\sqrt{2}+3.2+\sqrt{2^2}+\left(\sqrt{2}\right)^3\right]}+\sqrt[3]{\left[2^3-3.2.\sqrt{2}+3.2.\sqrt{2^2}-\left(\sqrt{2}\right)^3\right]}\)
\(=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2-\sqrt{2}\right)^3}\)
\(=2+\sqrt{2}+2-\sqrt{2}\)
\(=4\)
Vậy x = 4.