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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: \(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}+\dfrac{3}{17\cdot20}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)=\dfrac{1}{3}\cdot\dfrac{9}{20}=\dfrac{3}{20}\)
Ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{7}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2009.\frac{1}{7}=287\Rightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=287\)\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=287\)
\(\Rightarrow\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=287-3=284\)
\(\Rightarrow S=284\)
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
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!
\(=>2A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{101}}\)
\(=>2A-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{101}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{100}}\right)\)
\(=>A=\dfrac{1}{2^{101}}-\dfrac{1}{2}\)
\(A=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4031}{2015^2.2016^2}\)
\(A=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{2016^2-2015^2}{2015^2.2016^2}\)
\(A=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{2015^2}-\dfrac{1}{2016^2}\)
\(A=1-\dfrac{1}{2016^2}< 1\left(đpcm\right)\)
Bài 1:
\(\frac{1}{8}.16^n=2^n\)
\(\Rightarrow\frac{16^n}{8}=2^n\)
\(\Rightarrow\frac{\left(2^4\right)^n}{2^3}=2^n\)
\(\Rightarrow\frac{2^{4n}}{2^3}=2^n\)
\(\Rightarrow2^{4n-3}=2^n\)
\(\Rightarrow4n-3=n\)
\(\Rightarrow4n-n=3\)
\(\Rightarrow3n=3\)
\(\Rightarrow n=3:3\)
\(\Rightarrow n=1\left(TM\right).\)
Vậy \(n=1.\)
Bài 3:
a) \(\left|2x+3\right|=x+2\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-x=2-3\\2x+x=-2-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1x=-1\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\left(-1\right):1\\x=\left(-5\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{5}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-1;-\frac{5}{3}\right\}.\)
Chúc bạn học tốt!
Bài 3:
b) \(A=\left|x-2006\right|+\left|2007-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|x-2006\right|+\left|2007-x\right|\ge\left|x-2006+2007-x\right|\)
\(\Rightarrow A\ge\left|1\right|\)
\(\Rightarrow A\ge1.\)
Dấu '' = '' xảy ra khi:
\(\left(x-2006\right).\left(2007-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2006\ge0\\2007-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2006\le0\\2007-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2006\\x\le2007\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2006\\x\ge2007\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2006\le x\le2007\\x\in\varnothing\end{matrix}\right.\)
Vậy \(MIN_A=1\) khi \(2006\le x\le2007.\)
Chúc bạn học tốt!
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x;1-2y\in U\left(40\right)\)
\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
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
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