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\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (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
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.
a) Ta có: \(A=\sqrt{3+2\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
\(=\sqrt{1+2\cdot1\cdot\sqrt{2}+2}-\frac{1}{1+\sqrt{2}}\)
\(=\sqrt{\left(1+\sqrt{2}\right)^2}-\frac{1}{1+\sqrt{2}}\)
\(=1+\sqrt{2}-\frac{1}{1+\sqrt{2}}\)
\(=\frac{\left(1+\sqrt{2}\right)^2}{1+\sqrt{2}}-\frac{1}{1+\sqrt{2}}\)
\(=\frac{1+2\sqrt{2}+2-1}{1+\sqrt{2}}\)
\(=\frac{2\sqrt{2}+2}{1+\sqrt{2}}\)
\(=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=2\)
b) Ta có: \(\left(\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{3}{\sqrt{x}-3}\right)\cdot\frac{\sqrt{x}+3}{x+9}\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right)\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{1}{\sqrt{x}-3}\)
\(=\frac{1}{\sqrt{x}-3}\)(đpcm)
b. ĐK \(\hept{\begin{cases}x-2\ge0\\y+2014\ge0\\z-2015\ge o\end{cases}\Rightarrow\hept{\begin{cases}x\ge2\\y\ge-2014\\z\ge2015\end{cases}}}\)
Ta có \(\sqrt{x-2}+\sqrt{y+2014}+\sqrt{z-2015}=\frac{1}{2}\left(x+y+z\right)\)
Đặt \(\hept{\begin{cases}\sqrt{x-2}=a\ge0\\\sqrt{y+2014}=b\ge0\\\sqrt{z-2015}=c\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x-2=a^2\\y+2014=b^2\\z-2015=c^2\end{cases}\Rightarrow x+y+z}=a^2+b^2+c^2+3\)
Pt \(\Leftrightarrow a+b+c=\frac{1}{2}\left(a^2+b^2+c^2+3\right)\Leftrightarrow a^2+b^2+c^2+3=2a+2b+2c\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow\hept{\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}}\)\(\Leftrightarrow a=b=c=1\)
\(\Rightarrow\hept{\begin{cases}x-2=1\\y+2014=1\\z-2015=1\end{cases}\Rightarrow\hept{\begin{cases}x=3\\y=-2013\\z=2016\end{cases}\left(tm\right)}}\)
Vậy \(x=3;y=-2013;z=2016\)