\(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
K
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S
11 tháng 8 2019
\(\frac{x}{\sqrt{x}+\sqrt{y}}-\frac{y}{\sqrt{x}+\sqrt{y}}=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{x}-\sqrt{y}\)
\(tt:\frac{y-z}{\sqrt{y}+\sqrt{z}}=\sqrt{y}-\sqrt{z};.....\)
\(\Rightarrow\frac{x}{\sqrt{x}+\sqrt{y}}-\frac{y}{\sqrt{y}+\sqrt{x}}+.....-\frac{x}{\sqrt{x}+\sqrt{z}}=0\Rightarrow dpcm\)
AH
Akai Haruma
Giáo viên
27 tháng 5 2019
Lời giải:
ĐK: \(x,y\geq 0; x\neq y\). Để cho gọn đặt \(\sqrt{x}=a; \sqrt{y}=b\). Khi đó:
\(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}-y\sqrt{y}}{x-y}\right).\frac{(\sqrt{x}-\sqrt{y})^2}{x\sqrt{x}+y\sqrt{y}}\)
\(=(\frac{a^2-b^2}{a-b}-\frac{a^3-b^3}{a^2-b^2}).\frac{(a-b)^2}{a^3+b^3}\)
\(=\frac{(a^2-b^2)(a+b)-(a^3-b^3)}{a^2-b^2}.\frac{(a-b)^2}{a^3+b^3}\)
\(=\frac{ab(a-b)}{(a-b)(a+b)}.\frac{(a-b)^2}{a^3+b^3}=\frac{ab(a-b)^2}{(a+b)(a^3+b^3)}\)
\(=\frac{\sqrt{xy}(\sqrt{x}-\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})(x\sqrt{x}+y\sqrt{y})}\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
ĐKXĐ: ...
Lấy pt cuối trừ 3 lần pt đầu ta được:
\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^3+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^3+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^3=\frac{512}{27}\)
Pt (2) tương đương:
\(x+\frac{1}{x}-2+y+\frac{1}{y}-2+z+\frac{1}{z}-2=\frac{64}{9}\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=\frac{64}{9}\)
Đặt \(\left(\sqrt{x}-\frac{1}{\sqrt{x}};\sqrt{y}-\frac{1}{\sqrt{y}};\sqrt{z}-\frac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\)
Hệ trở thành:
\(\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\a^2+b^2+c^2=\frac{64}{9}\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\)
Ta có: \(a^3+b^3+c^3-3abc=\frac{512}{27}-3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=\frac{512}{27}-3abc\)
\(\Leftrightarrow\frac{8}{3}.\left(\frac{64}{9}-0\right)=\frac{512}{27}-3abc\)
\(\Rightarrow abc=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\abc=0\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;0;\frac{8}{3}\right)\) và hoán vị
Hay \(\left(x;y;z\right)=\left(1;1;9\right)\) và hoán vị
11 tháng 7 2017
\(=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\right):\frac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(=\left[\left(\sqrt{x}+\sqrt{y}\right)-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right].\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)
\(=\frac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}.\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}=\frac{\sqrt{xy}}{x-\sqrt{xy}+y}\)
16 tháng 10 2019
\(\left(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\cdot\frac{1}{x-y}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\cdot\frac{1}{x-y}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(x-\sqrt{xy}+y-\sqrt{xy}\right)\cdot\frac{1}{x-y}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(x-2\sqrt{xy}+y\right)\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}+\frac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\sqrt{x}-\sqrt{y}+2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
\(=1\)