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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.
1.
= -(13 + 3 căn7 ) / 2 + -(7 + 3 căn7 ) / 2
= -7 + 3 căn7
Bài 2
\(P=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{4-\sqrt{12}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\left(\sqrt{3}+1\right)}\)
=\(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}+1\right)}=1\)
Vậy P là một số nguyên
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
\(A=\sqrt[3]{2^3+3.2^2.\sqrt{2}+3.2.\sqrt{2}^2+\sqrt{2}^3}+\sqrt[3]{\sqrt{2}^3-3.\sqrt{2}^2.2+3.\sqrt{2}.2^2-2^3}\)
\(A=\sqrt[3]{\left(2+\sqrt{2}\right)^3}+\sqrt[3]{\left(\sqrt{2}-2\right)^3}\)
\(A=2+\sqrt{2}+\sqrt{2}-2=2\sqrt{2}\)
\(X=\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\)
\(\Rightarrow X^3=\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)^3\)
\(\Rightarrow X^3=2+3\sqrt[3]{1-\frac{84}{81}}\left(\sqrt[3]{1+\frac{\sqrt{84}}{9}}+\sqrt[3]{1-\frac{\sqrt{84}}{9}}\right)\)
\(\Rightarrow X^3=2-3\sqrt[3]{\frac{1}{27}}.X\)
\(\Rightarrow X^3=2-X\)
\(\Rightarrow X^3+X-2=0\)
\(\Rightarrow\left(X-1\right)\left(X^2+2X+2\right)=0\)
\(\Rightarrow X=1\) (do \(X^2+2X+2=\left(X+1\right)^2+1>0\) \(\forall X\))