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Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
a: \(A\cdot B=\left(9-3\sqrt{7}\right)\left(9-3\sqrt{7}\right)=\left(9-3\sqrt{7}\right)^2=144-54\sqrt{7}\)
\(A+B=9-3\sqrt{7}+9-3\sqrt{7}=18-6\sqrt{7}\)
c: \(P=\dfrac{1+\sqrt{x}-1+\sqrt{x}}{1-x}:\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{-\left(x-1\right)}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-1}=-\dfrac{2x}{\left(x-1\right)\left(\sqrt{x}-1\right)}< 0\)
\(1.A=\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right).\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\left(\dfrac{3+\sqrt{5}}{9-5}-\dfrac{3-\sqrt{5}}{9-5}\right).\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{2\sqrt{5}}{4}.\sqrt{5}=\dfrac{5}{2}\) \(2.B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{100}-1\)
\(3.C=\sqrt[3]{7+5\sqrt{2}}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.2.1+3.\sqrt{2}.1+1}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.2.1+3.\sqrt{2}.1-1}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\) \(4.Sai-đề\) ???
Sorry và cám ơn bạn.
4.\(\sqrt[3]{9+4\sqrt{5}}\) + \(\sqrt[3]{9-4\sqrt{5}}\)