Cho \(\frac{x^{\text{4}}}{a}+\frac{y^{\text{4}}}{b}=\frac{1}{a+b};x^2+y^2=1\)
Chứng minh rằng:\(\frac{x^{200\text{4}}}{a^{1002}}+\frac{y^{200\text{4}}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
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2 tháng 9 2016
ĐKXĐ : \(x,y>0\)
a/ \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}+\frac{x+y}{\sqrt{xy}}\right)\)
\(=\left(\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right).\sqrt{x}}-\frac{y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}.\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{-\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)
\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{-\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{x+y}=\sqrt{y}-\sqrt{x}\)
b/ Ta có ; \(4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow B=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)
21 tháng 5 2015
\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)
\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)
\(A=\frac{3}{2}\times\frac{8}{99}\)
\(A=\frac{4}{33}\)
b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)
\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)
c] Ta đặt \(\left(8n+5,6n+4\right)=d\)
\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)
Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản
Ta có:
\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)
Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)
\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)
\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)
\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)
\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)
\(\Leftrightarrow x^2b-y^2a=0\)
\(\Leftrightarrow x^2b=y^2a\)
\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)
\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)
Chúc bạn học tốt!