Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))
Đặt \(A=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow A^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3.\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)^2\left(1-\frac{\sqrt{84}}{9}\right)}+3.\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)^2}\)
\(A^3=2+3.\sqrt[3]{-\frac{1}{27}.\left(1+\frac{\sqrt{84}}{9}\right)}+3.\sqrt[3]{-\frac{1}{27}.\left(1-\frac{\sqrt{84}}{9}\right)}\)
\(=2-\left(\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)}+\sqrt[.3]{\left(1-\frac{\sqrt{84}}{9}\right)}\right)\)
\(A^3=2-A\Leftrightarrow\left(A-1\right)\left(A^2+A+2\right)=0\Rightarrow A=1\)
Đặt \(A=\sqrt[3]{\frac{9+2\sqrt{21}}{9}}+\sqrt[3]{\frac{9-2\sqrt{21}}{9}}\)
\(A^3=\frac{9+2\sqrt{21}+9-2\sqrt{21}}{9}+3\sqrt[3]{\frac{9^2-4\cdot21}{9^2}}A\)
\(A^3-2+A=0\Leftrightarrow\left(A-1\right)\left(A^2+A+1\right)+A-1=0\Leftrightarrow\left(A-1\right)\left(A^2+A+2\right)=0\)
\(\Rightarrow A=1\)(ĐPCM)
Chứng minh rằng \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\) là một số nguyên
Đặt \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}=a;\sqrt[3]{1-\frac{\sqrt{84}}{9}}=b\Rightarrow x=a+b;a^3+b^3=2;ab=-\frac{1}{3}\)
Ta có:\(x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow x^3=2-x\Leftrightarrow x^3+x-2=0\Leftrightarrow\left(x-1\right).\left(x^2+x+2\right)=0\)
\(\Leftrightarrow x=1\).Vì \(x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)
Từ đó suy ra điều phải chứng minh
~~~~~~~~~~~ Chúc bạn hok tốt~~~~~~~~~~~~
Tính giá trị của biểu thức \(P=x^3+y^3-3\left(x+y\right)+2004\)
biết \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)và \(y=\sqrt[3]{17+12\sqrt{2}}+\sqrt[3]{17-12\sqrt{2}}\)
Chứng minh rằng \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\) là một số nguyên
Đặt \(\sqrt[3]{1+\frac{\sqrt{84}}{9}}=a\);\(\sqrt[3]{1-\frac{\sqrt{84}}{9}}=b\)
\(\Rightarrow x=a+b;a^3+b^3=2;ab=-\frac{1}{3}\)
Ta có: \(x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow x^3=2-x\Leftrightarrow x^3+x-2=0\Leftrightarrow\left(x-1\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow x=1\).vì \(x^2+x+2=0=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\)
=> đpcm
P/s tham khảo
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
Đặt \(P=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(P^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3\sqrt[3]{\left(1+\frac{\sqrt{84}}{9}\right)\left(1-\frac{\sqrt{84}}{9}\right)}\cdot P\)
\(P^3=2+3\sqrt[3]{1-\frac{84}{81}}\cdot P\)
\(P^3=2+3\sqrt[3]{\frac{-1}{27}}\cdot P\)
\(P^3=2+3\cdot\frac{-1}{3}\cdot P\)
\(P^3=2-P\)
\(\Leftrightarrow P^3+P-2=0\)
\(\Leftrightarrow P^3-P^2+P^2-P+2P-2=0\)
\(\Leftrightarrow P^2\left(P-1\right)+P\left(P-1\right)+2\left(P-1\right)=0\)
\(\Leftrightarrow\left(P-1\right)\left(P^2+P+2\right)=0\)
Do \(P^2+P+2>0\forall P\)
Do đó \(P-1=0\Leftrightarrow P=1\)
Vậy \(P=1\) là một số nguyên ( đpcm )
Ta có : \(x=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Leftrightarrow x^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Leftrightarrow x^3=1+\frac{\sqrt{84}}{9}+1-\frac{\sqrt{84}}{9}+3.\sqrt[3]{1+\frac{\sqrt{84}}{9}}.\sqrt[3]{1-\frac{\sqrt{84}}{9}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}^3\right)\)
\(\Leftrightarrow x^3=2+3.\sqrt[3]{1^2-\frac{84}{81}}.x\Leftrightarrow x^3=2-x\)
\(\Leftrightarrow x^3+x-2=0\Leftrightarrow\left(x-1\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x^2+x+2=0\end{array}\right.\)
Vì \(x^2+x+2=\left(x^2+x+\frac{1}{4}\right)+\frac{7}{4}=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\) nên pt này vô nghiệm.
Vậy x - 1 = 0 => x = 1
Vậy x có giá trị là số nguyên.