so sanh x va y biet
a) x=\(2\sqrt{7}\)va y=\(3\sqrt{3}\)
b) x=\(6\sqrt{2}\)va y=\(5\sqrt{3}\)
c) x=\(\sqrt{31}-\sqrt{33}\) va y=\(6-\sqrt{11}\)
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1: \(\left(\sqrt{3}+\sqrt{7}\right)^2=10+2\sqrt{21}\)
\(\left(2+\sqrt{6}\right)^2=10+4\sqrt{6}\)
mà 2 căn 21<4 căn 6
nên căn 3+căn 7<2+căn 6
2: \(\sqrt{7}-\sqrt{5}=\dfrac{2}{\sqrt{7}+\sqrt{5}}\)
\(\sqrt{6}-2=\dfrac{2}{\sqrt{6}+2}\)
mà \(\sqrt{7}+\sqrt{5}>\sqrt{6}+2\)
nên \(\sqrt{7}-\sqrt{5}< \sqrt{6}-2\)
3: \(\sqrt{11}-\sqrt{7}=\dfrac{4}{\sqrt{11}+\sqrt{7}}\)
\(\sqrt{7}-\sqrt{3}=\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
mà căn 11>căn 3
nên \(\sqrt{11}-\sqrt{7}< \sqrt{7}-\sqrt{3}\)
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
Ta có \(ax^3=by^3=cz^3\Leftrightarrow\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=ax^2+by^2+cz^2\Leftrightarrow\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}=\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}+\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}+\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)Vậy \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
a: \(x=2\sqrt{7}=\sqrt{28}>\sqrt{27}=y\)
b: \(x=6\sqrt{2}=\sqrt{72}< \sqrt{75}=y\)
Giả sử \(\sqrt{2009}\ge2\sqrt{2008}-\sqrt{2007}\)
\(\Leftrightarrow\sqrt{2009}-\sqrt{2008}\ge\sqrt{2008}-\sqrt{2007}\)
\(\Leftrightarrow\frac{1}{\sqrt{2009}+\sqrt{2008}}\ge\frac{1}{\sqrt{2008}+\sqrt{2007}}\) (sai)
Vậy \(\sqrt{2009}< 2\sqrt{2008}-\sqrt{2007}\)
a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)