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Ta có :
\(\sqrt{50}+\sqrt{5}>\sqrt{49}+\sqrt{4}=7+2=9\)
Vậy \(\sqrt{50}+\sqrt{5}>9\)
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Đặt \(\sqrt{2011}=a;\sqrt{2012}=b\)
Theo đề, ta có: \(A=\dfrac{a^2}{b}+\dfrac{b^2}{a}=\dfrac{a^3+b^3}{ab}\)
B=a+b
\(A-B=\dfrac{a^3+b^3}{ab}-\left(a+b\right)=\dfrac{a^3+b^3-a^2b-ab^2}{ab}\)
\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)}{ab}\)
\(=\dfrac{\left(a+b\right)\left(a-b\right)^2}{ab}>0\)
=>A>B
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Ta có: \(\sqrt{2}>1\)
\(\Rightarrow1+\sqrt{2}>1+1\)
\(\Rightarrow1+\sqrt{2}>2\)
Ta có:\(\sqrt{2}>\sqrt{1}\)
\(\Leftrightarrow1+\sqrt{2}>1+\sqrt{1}=2\)
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a)\(10^{20}=\left(10^2\right)^{10}=100^{10}\left(1\right)\)
\(9^{30}=\left(9^3\right)^{10}=729^{10}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow9^{30}>10^{20}\)
b) \(\left(-5\right)^{30}=5^{30}=125^{10}\)
\(\left(-3\right)^{50}=3^{50}=243^{10}\)
\(\Rightarrow\left(-3\right)^{50}>\left(-5\right)^{30}\)
c)\(64^8=\left(2^6\right)^8=2^{48}\)
\(16^{12}=\left(2^4\right)^{12}=2^{48}\)
\(\Rightarrow64^8=16^{12}\)
\(\sqrt{8}\)-\(\sqrt{5}\)
<1
Ta có : \(1=3-2=\sqrt{9}-\sqrt{4}\)
Vì \(\left\{{}\begin{matrix}\sqrt{9}>\sqrt{8}\\\sqrt{4}< \sqrt{5}\end{matrix}\right.\Rightarrow}\left\{{}\sqrt{8}-\sqrt{5}< \sqrt{9}-\sqrt{4}=1}\)