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\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4031}{2015^2.2016^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{2016^2-2015^2}{2015^2.2016^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)
\(A=1-\dfrac{1}{2016^2}< 1\left(đpcm\right)\)
a,|x2−13x2−13| = 3232
b, 32−1232−12 ( 2x-1)=3434
c, |x-1|+2x=2
a)\(\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=\dfrac{3}{2}\)
TH1
\(\dfrac{x}{2}-\dfrac{1}{3}=\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=\dfrac{11}{6}\)
=>x=\(\dfrac{11.2}{6}\)
=>x=\(\dfrac{11}{3}\)
TH2
\(\dfrac{x}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=-\dfrac{3}{2}+\dfrac{1}{2}\)
=>\(\dfrac{x}{2}=-1\)
=>x=-2
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x;1-2y\in U\left(40\right)\)
\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Mà 1-2y lẻ nên:
\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)
b tương tự.
c) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)
d tương tự
\(=>2A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{101}}\)
\(=>2A-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{101}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{100}}\right)\)
\(=>A=\dfrac{1}{2^{101}}-\dfrac{1}{2}\)
\(\sqrt{\dfrac{16}{169}}.\dfrac{-3}{2}.\left(\dfrac{3}{2}+\dfrac{-5}{12}\right):\left(-\dfrac{1}{2}\right)\\ =\left|\sqrt{\left(\pm\dfrac{4}{13}\right)^2}\right|.\dfrac{-3}{2}.\dfrac{13}{12}.\left(-2\right)\\ =\left(\dfrac{4}{13}.\dfrac{13}{12}\right).\left(-2.\dfrac{-3}{2}\right)\\ =\dfrac{4}{12}.3=\dfrac{12}{12}=1\)
a)
ĐKXĐ: \(2x\geq 0\Leftrightarrow x\geq 0\)
Vậy TXĐ của $x$ là \(D= [0;+\infty)\)
b)
ĐK: \((2x-1)(x+3)\neq 0\Leftrightarrow \left\{\begin{matrix} 2x-1\neq 0\\ x+3\neq 0\end{matrix}\right.\Leftrightarrow \Leftrightarrow \left\{\begin{matrix} x\neq \frac{1}{2}\\ x\neq -3\end{matrix}\right.\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{1}{2}; -3\right\}\)
c)
ĐK: \(8x^3+1\neq 0\Leftrightarrow x^3\neq \frac{-1}{8}\Leftrightarrow x\neq \frac{-1}{2}\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{-1}{2}\right\}\)
d)
ĐK:
\(|x-2015|+1\neq 0\Leftrightarrow |x-2015|\neq -1\Leftrightarrow x\in\mathbb{R}\)
Vậy TXĐ \(D=\mathbb{R}\)
e)
ĐK: \(\left\{\begin{matrix} |x-1,2|\neq 0\\ 2x-5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 1,2\\ x\neq 2,5\end{matrix}\right.\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{1,2; 2,5\right\}\)
f)
ĐK: \(x^2-4\neq 0\Leftrightarrow (x-2)(x+2)\neq 0\Leftrightarrow x\neq \pm 2\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{\pm 2\right\}\)
a: \(=\dfrac{2}{3}\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{60}-\dfrac{1}{63}+...+\dfrac{1}{117}-\dfrac{1}{120}\right)+\dfrac{2}{2006}\)
\(=\dfrac{2}{3}\cdot\dfrac{1}{120}+\dfrac{1}{2003}=\dfrac{1}{180}+\dfrac{1}{2003}=\dfrac{2183}{180\cdot2003}\)
b: \(=\dfrac{5}{4}\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\left(\dfrac{1}{40}-\dfrac{1}{80}\right)+\dfrac{5}{2006}\)
\(=\dfrac{5}{4}\cdot\dfrac{1}{80}+\dfrac{5}{2006}=\dfrac{1}{64}+\dfrac{5}{2006}=\dfrac{1163}{64192}\)
c: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)
Đặt \(A=\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{2019^3}\)
\(\Rightarrow2A=\frac{2}{2^3}+\frac{2}{3^3}+...+\frac{2}{2019^3}\)
Ta có:
\(\left\{{}\begin{matrix}\frac{2}{2^3}< \frac{2}{1.2.3}\\\frac{2}{3^3}< \frac{1}{2.3.4}\\....\\\frac{2}{2019^3}< \frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\end{matrix}\right.\)
\(\Rightarrow2A< \frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{\left(2019-1\right).2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(2019-1\right).2019}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.\left(2019+1\right)}\)
\(\Rightarrow2A< \frac{1}{1.2}-\frac{1}{2019.2020}\)
\(\Rightarrow A< \left(\frac{1}{1.2}-\frac{1}{4078380}\right):2\)
\(\Rightarrow A< \frac{1}{1.2}:2-\frac{1}{4078380}:2\)
\(\Rightarrow A< \frac{1}{4}-\frac{1}{8156760}\)
\(\Rightarrow A< \frac{1}{2^2}-\frac{1}{8156760}\)
Vì \(\frac{1}{2^2}-\frac{1}{8156760}< \frac{1}{2^2}.\)
\(\Rightarrow A< \frac{1}{2^2}\left(đpcm\right).\)
Chúc bạn học tốt!