\(\frac{\sqrt[3]{20+14\sqrt{2}}}{\sqrt[3]{20-14\sqrt{2}}}\)
Những câu hỏi liên quan
a, A = \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{14\sqrt{2}-20}\)
b, X = \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))
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2 tháng 8 2021 lúc 16:51
\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
Tính
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 \(\sqrt[3]{20+14\sqrt[]{2}}+\sqrt[3]{20-14\sqrt[]{2}}=4\)
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Chứng minh rằng các biểu thức sau là 1 số nguyên:
a) \(A=\sqrt[3]{20+14\sqrt{2}}-\sqrt[3]{14\sqrt{2}-20}\)
b) \(B=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
rut gon $\sqrt[3]{20+14$\sqrt{2}$}$ + $\sqrt[3]{20-14$\sqrt{2}$}$
\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
= \(\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(2+\sqrt{2}\right)^3}\) = \(2+\sqrt{2}+2+\sqrt{2}\) = 4+\(2\sqrt{2}\)
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Lộn nha = \(2\sqrt[3]{20+14\sqrt{2}}\) mới đúng nha.
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Xem thêm câu trả lời
\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
A = \(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
=> A3 = 40 + 6A
<=> A = 4
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Rút gọn biểu thức :
a) A=\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\).
b)B=\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
c) C=\(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}.\)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
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Tính các giá trị của\(A=x^3-6x\) tại \(x=\sqrt[3]{14\sqrt{2}+20}+\sqrt[3]{-14\sqrt{2}+20}\)
`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`
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1. Cho x=\(\sqrt[3]{7+5\sqrt{2}}-\frac{1}{\sqrt[3]{7-5\sqrt{2}}}\)
Chứng minh rằng: \(^{x^3+3x-14=0}\)
2. Cho x=\(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\)
Tính: A=\(\left(x^3-3x^2+x-19\right)^{2019}\)