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a) \(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}\)
\(=\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^2\left(n^2+2n+1+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{n^4+2n^2\left(n+1\right)+\left(n+1\right)^2}{n^2\left(n+1\right)^2}\)
\(=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=>đpcm
b) Từ công thức trên ta có:
\(1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}=\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}\)
=> \(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(n+1\right)}=1+\frac{1}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Ta có:
\(S=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(=2010+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2010}-\frac{1}{2011}\right)\)
\(2010+\left(1-\frac{1}{2011}\right)=2010+\frac{2010}{2011}=2010\frac{2010}{2011}\)
\(A=\frac{1}{\sqrt{2.1}\left(\sqrt{2}+\sqrt{1}\right)}+\frac{1}{\sqrt{2.3}\left(\sqrt{3}+\sqrt{2}\right)}+\frac{1}{\sqrt{3.4}\left(\sqrt{4}+\sqrt{3}\right)}+...+\frac{1}{\sqrt{999.1000}\left(\sqrt{1000}+\sqrt{999}\right)}\)
\(A=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{2.1}\left(2-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2.3}\left(3-2\right)}+\frac{\sqrt{4}-\sqrt{3}}{\sqrt{3.4}\left(4-3\right)}+...+\frac{\sqrt{1000}-\sqrt{999}}{\sqrt{999.1000}\left(1000-999\right)}\)
\(A=\frac{\sqrt{2}}{\sqrt{2.1}}-\frac{\sqrt{1}}{\sqrt{2.1}}+\frac{\sqrt{3}}{\sqrt{2.3}}-\frac{\sqrt{2}}{\sqrt{2.3}}+\frac{\sqrt{4}}{\sqrt{3.4}}-\frac{\sqrt{3}}{\sqrt{3.4}}+...+\frac{\sqrt{1000}}{\sqrt{999.1000}}-\frac{\sqrt{999}}{\sqrt{1000.999}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{999}}-\frac{1}{\sqrt{1000}}\)
\(A=\frac{1}{1}-\frac{1}{\sqrt{1000}}=\frac{\sqrt{1000}-1}{\sqrt{1000}}=\frac{10\sqrt{10}-1}{10\sqrt{10}}\)
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(B=\frac{1}{4}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1}{4}\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
4) mấy bài kia trình bày dài lắm!! (lười ý mà ahihi)
\(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+|x+y+z|=0.\)
\(\Leftrightarrow|x-\sqrt{2}|+|y+\sqrt{2}|+|x+y+z|=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{2}=0\\y+\sqrt{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}\\y=-\sqrt{2}\end{cases}}}\)
Tìm z thì dễ rồi
d: \(D=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4:\dfrac{13}{12}\)
\(=-4\cdot\dfrac{12}{13}=-\dfrac{48}{13}\)
e: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
=20-12+5-6
=8+5-6
=13-6=7
f: \(F=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{2}:6\)
\(=\dfrac{9}{4}+\dfrac{5}{6}-\dfrac{3}{12}\)
\(=\dfrac{27}{12}+\dfrac{10}{12}-\dfrac{3}{12}=\dfrac{34}{12}=\dfrac{17}{6}\)
\(M=4\frac{1}{3}-\sqrt{16}+5\sqrt{\frac{4}{9}}-\frac{25}{\left(\sqrt{6}\right)^2}\)
\(=\frac{13}{3}-4+5\cdot\frac{2}{3}-\frac{25}{6}\)
\(=\frac{1}{3}+\frac{10}{3}-\frac{25}{6}\)
\(=\frac{11}{3}-\frac{25}{6}\)
\(=-\frac{1}{2}\)