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5 tháng 7 2016
Xét : Với mọi \(x\in N^{\text{*}}\) , ta có : \(\frac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\frac{1}{\sqrt{x\left(x+1\right)}\left(\sqrt{x}+\sqrt{x+1}\right)}=\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x\left(x+1\right)}}=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\)
Áp dụng vào tính : \(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
NB
6 tháng 3 2022
\(=\left(4\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{12+4\sqrt{6}+2+8\sqrt{3}+4\sqrt{2}+4-2}\right)\\ =\left(4\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{\left(2\sqrt{3}+\sqrt{2}\right)^2+4\left(2\sqrt{3}+\sqrt{2}\right)+4-2}\right)\\ =\left(4\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{\left(2\sqrt{3}+\sqrt{2}+2\right)^2-2}\right)\\ =\left(4\sqrt{3}-2\sqrt{2}\right)\left(2\sqrt{3}+\sqrt{2}\right)=20\)
28 tháng 9 2021
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\\ \Leftrightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{81-80}=18-3x\\ \Leftrightarrow x^3-3x=18\\ y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\\ \Leftrightarrow y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{9-8}=6+3y\\ \Leftrightarrow y^3-3y=6\\ \Leftrightarrow P=x^3+y^3-3\left(x+y\right)+1993\\ P=x^3+y^3-3x-3y+1993=18+6+1993=2017\)
28 tháng 9 2021
Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}.x=18+3x\)
\(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\)
\(\Rightarrow y^3=3-2\sqrt{2}+3+2\sqrt{2}+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}y=6+3y\)
\(P=x^3+y^3-3\left(x+y\right)+1993\)
\(=18+3x+6+3y-3x-3y+1993=2017\)
6 tháng 11 2021
a, \(\Rightarrow M=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(\Rightarrow M=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b, \(x=3+2\sqrt{2}\Rightarrow M=\dfrac{\sqrt{3+2\sqrt{2}}-2}{\sqrt{3+2\sqrt{2}}}=\dfrac{\sqrt{2+2\sqrt{2}.1+1}-2}{\sqrt{2+2\sqrt{2}.1+1}}=\dfrac{\sqrt{2}+1-2}{\sqrt{2}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}=\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\dfrac{2-2\sqrt{2}+1}{2-1}=3-2\sqrt{2}\)
c, \(M>0\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\Rightarrow\sqrt{x}-2>0\Rightarrow\sqrt{x}>2\Rightarrow x>4\)
12 tháng 9 2021
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
12 tháng 9 2021
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
10 tháng 12 2021
\(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{\left(n+1\right)^2}}\\ =\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+\dfrac{2}{n}-\dfrac{2}{n+1}-\dfrac{2}{n\left(n+1\right)}}\\ =\sqrt{\left[1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right]^2}=\left|1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right|\)
\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)