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\(a_n=\frac{2}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(n+1-n\right)}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+n+1}\)
\(< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(a_1+a_2+a_3+...+a_{2009}< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...-\frac{1}{\sqrt{2010}}=1-\frac{1}{\sqrt{2010}}< \frac{2008}{2010}\)
1. Ta có: \(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Rightarrow\left(x+y+z\right)^2=\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz=xy+yz+zx+2y\sqrt{xz}+2z\sqrt{xy}+2x\sqrt{yz}\)
\(\Leftrightarrow x^2+y^2+z^2+xy+yz+zx-2y\sqrt{xz}-2z\sqrt{xy}-2x\sqrt{yz}=0\)
\(\Leftrightarrow\left(x-\sqrt{yz}\right)^2+\left(y-\sqrt{xz}\right)^2+\left(z-\sqrt{xy}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{yz}\\y=\sqrt{xz}\\z=\sqrt{xy}\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\Rightarrow x=y=z\)
Bài 1:
\(x+y+z=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
\(\Leftrightarrow x+y+z-\sqrt{xy}-\sqrt{yz}-\sqrt{xz}=0\)
\(\Leftrightarrow 2x+2y+2z-2\sqrt{xy}-2\sqrt{yz}-2\sqrt{xz}=0\)
\(\Leftrightarrow (x+y-2\sqrt{xy})+(y+z-2\sqrt{yz})+(z+x-2\sqrt{xz})=0\)
\(\Leftrightarrow (\sqrt{x}-\sqrt{y})^2+(\sqrt{y}-\sqrt{z})^2+(\sqrt{z}-\sqrt{x})^2=0\)
Vì \( (\sqrt{x}-\sqrt{y})^2;(\sqrt{y}-\sqrt{z})^2;(\sqrt{z}-\sqrt{x})^2\geq 0, \forall x,y,z>0\) nên để tổng của chúng bằng $0$ thì:
\( (\sqrt{x}-\sqrt{y})^2=(\sqrt{y}-\sqrt{z})^2=(\sqrt{z}-\sqrt{x})^2=0\)
\(\Rightarrow x=y=z\) (đpcm)
1. Với mọi số thực x;y;z ta có:
\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)
\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)
\(\Rightarrow P\ge3\)
\(P_{min}=3\) khi \(x=y=z=1\)
1.1
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)
\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)
\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)
\(\Leftrightarrow a=b\Leftrightarrow x=y\)
Thay vào pt đầu:
\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))
\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)
\(\Rightarrow a=1\Rightarrow x=y=1\)
2.
\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)
\(\Rightarrow4x^2-10xy+4y^2=0\)
\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu
...
1 + y2 = xy + yz + xz + y2 = (x + y)(y + z)
1 + z2 = xy + yz + xz + z2 = (x + z)(z + y)
1 + x2 = xy + yz + xz + x2 = (y + x)(x + z)
Sau khi thay vào và rút gọn ta được
S = x(y + z) + y(x + z) + z(x + y)
S = 2(xy + yz + xz) = 2.1 = 2
