\(y=\frac{1}{x^2+\sqrt{x}}\int^{ }_{ }^{ }^2_{ }_{ }\vec{\vec{\vec{^2\frac{\frac{\frac{\sqrt[]{}\sqrt[]{}\sqrt[]{}\sqrt[]{}\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{\frac{ }{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}{ }}}}\)
\(y=\frac{1}{x^2+\sqrt{x}}\Delta\theta\lambda\vec{\cos^2\int^{ }_{ }\frac{\sqrt[]{}\sqrt{ }}{ }}\)
\(y=\frac{X+1}{2014}\vec{+\vec{\frac{X+2}{2013}\vec{=\vec{\vec{\frac{X+3}{2012}\vec{+\vec{\frac{X+4}{2011}}}}}}}}\) Giải phương trình sau :
\(y=\frac{1}{x^2+\sqrt{x}}\int^{ }_{ }^2^{ }^2_{ }\frac{\sqrt[]{}\sqrt{ }}{ }\)
\(x+y=4\Rightarrow\frac{x+y}{2}=2\Rightarrow\sqrt{\frac{x+y}{2}}=\sqrt{2}\)
\(P.\sqrt{\frac{x+y}{2}}=\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{2}\sqrt{x^2+\frac{1}{x^2}}\)
\(\Leftrightarrow\sqrt{2}P=\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}+\sqrt{1+1}\sqrt{x^2+\frac{1}{x^2}}\)
\(\Leftrightarrow\sqrt{2}P\ge x+\frac{1}{x}+y+\frac{1}{y}\)
\(x+\frac{1}{x}=\left(\frac{1}{x}+4x\right)-3x\ge4-3x\)
\(y+\frac{1}{y}=\left(\frac{1}{y}+4y\right)-3y\ge4-3y\)
\(\Rightarrow\sqrt{2}P\ge8-3\left(x+y\right)=8-3.4=-4\)
đến đay sau răng
Cho x;y;z>0;\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\) . CMR:\(\frac{\sqrt{x^2+2y^2}}{xy}+\frac{\sqrt{y^2+2z^2}}{yz}+\frac{\sqrt{z^2+2x^2}}{zx}\ge\sqrt{3}\)
Cho x;y;z>0. Chứng minh rằng: \(\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+y^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)\(\frac{1}{z^2}\)
\(y=1-\frac{\sqrt{x}}{\sqrt{x+}1}:\frac{2+\sqrt{x}}{x-5\sqrt{x+6}}+\frac{3+\sqrt{x}}{\sqrt{x-2}}-\frac{2+\sqrt{x}}{\sqrt{x-3}}\)
tính:\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
Tính :
a ) \(S=\frac{1}{\sqrt{1}\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+.....+\)\(\frac{1}{\sqrt{2017}+\sqrt{2019}}\)
b ) \(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+....+\frac{1}{\sqrt{100}+\sqrt{102}}\)
c ) \(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{101}}\)
d ) \(S=\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}}+....+\frac{1}{\sqrt{2016}+\sqrt{2019}}\)
