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\(x+2x+3x+...+2011x=2012.1013\)
\(\dfrac{2011\left(2011+1\right)}{2}x=2012.2013\)
\(x=2012.2013.\dfrac{2}{2011.2012}\)
\(x=\dfrac{4026}{2011}\)
Giải:
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Leftrightarrow\dfrac{x+4}{2008}+\dfrac{x+3}{2009}+2=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}+2\)
\(\Leftrightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Leftrightarrow\dfrac{x+4+2008}{2008}+\dfrac{x+3+2009}{2009}=\dfrac{x+2+2010}{2010}+\dfrac{x+1+2011}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Leftrightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên \(x+2012=0\)
\(\Leftrightarrow x=0-2012\)
\(\Leftrightarrow x=-2012\)
Vậy \(x=-2012\).
Chúc bạn học tốt!
\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)
\(\Rightarrow\dfrac{x+4}{2008}+1+\dfrac{x+3}{2009}+1=\dfrac{x+2}{2010}+1+\dfrac{x+1}{2011}+1\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\)
\(\Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)
\(\Rightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)
Vì \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\)
Nên:
\(x+2012=0\Rightarrow x=-2012\)
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2011}+\dfrac{x-2}{2012}+\dfrac{x-3}{2009}-\dfrac{x-4}{2008}=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)+\left(\dfrac{x-3}{2009}-1\right)+\left(\dfrac{x-4}{2008}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
Mà \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\)
\(\Leftrightarrow x-2012=0\Leftrightarrow x=2012\)
Vậy ...
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}+\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
=> \(\dfrac{x-1}{2011}-1+\dfrac{x-2}{2010}-1+\dfrac{x-3}{2009}-1=\dfrac{x-4}{2008}-1-2\)
=>\(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}=\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}\)
=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}+\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}-\dfrac{x-2012}{\left(x-2012\right):2}=0\)=> x - 2012 ( \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\)) = 0
Vì \(\dfrac{1}{2011}+\dfrac{1}{2010}+\dfrac{1}{2009}-\dfrac{1}{2008}-\dfrac{1}{\left(x-2012\right):2}\) \(\ge\) 0
=> x - 2012 = 0
=> x = 2012
Bài 1:
Ta có:
\(\left(\dfrac{ab}{2}-\dfrac{6ab}{7}\right):\dfrac{5b^2}{14}=\left(\dfrac{7ab}{14}-\dfrac{12ab}{14}\right).\dfrac{14}{5b^2}\)
\(=\dfrac{-5ab}{14}.\dfrac{14}{5b^2}=\dfrac{-a}{b}\)(1)
Thay \(a=\dfrac{2007}{2010};b=\dfrac{2011}{2010}\) vào (1) ta được:
\(\dfrac{-\dfrac{2007}{2010}}{\dfrac{2011}{2010}}=-\dfrac{2007}{2011}\)
Vậy......................
Chúc bạn học tốt!!!
Bài 2:
\(\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -\dfrac{1}{2}.\dfrac{3}{4}:\dfrac{1}{8}+1\)
\(\Rightarrow2.\left(-\dfrac{9}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -3+1\)
\(\Rightarrow-\dfrac{37}{4}< \dfrac{x}{8}< -2\)
\(\Rightarrow\dfrac{-74}{8}< \dfrac{x}{8}< -\dfrac{16}{8}\)
\(\Rightarrow-74< x< -16\)
\(\Rightarrow x\in\left\{-73;-72;-71;....;-18;-17\right\}\)
Vậy..............................
Chúc bạn học tốt!!!
a: \(\Leftrightarrow-\dfrac{3}{2x-3}=\dfrac{2}{5}-\dfrac{3}{2}-3=\dfrac{-41}{10}\)
=>41(2x-3)=30
=>82x-123=30
=>82x=153
hay x=153/82
b: \(\Leftrightarrow\left(x-1\right)\left(7-2x\right)=0\)
=>x=1 hoặc x=7/2
c: \(\Leftrightarrow\left(\dfrac{x+1}{2018}+1\right)+\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+10}{2009}+1\right)+\left(\dfrac{x+11}{2008}+1\right)+\left(\dfrac{x+12}{2007}+1\right)\)
=>x+2019=0
hay x=-2019
a) \(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\) \(\Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=-2\\\dfrac{x}{2}-\dfrac{1}{3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{-5}{3}\\\dfrac{x}{2}=\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{14}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{-10}{3}\) hoặc \(x=\dfrac{14}{3}\) thì thỏa mãn đề bài.
b) \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\) \(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\) \(\Rightarrow\dfrac{x+4+2010}{2010}+\dfrac{x+3+2011}{2011}=\dfrac{x+2+2012}{2012}+\dfrac{x+1+2013}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\) \(\Rightarrow\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\) \(\Rightarrow x+2014=0\) \(\Rightarrow x=-2014\)
Vậy \(x=-2014\) thì thỏa mãn đề bài.
c) \(3^{x+2}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1+1}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}\times3+4\times3^{x+1}=7\times3^6\) \(\Rightarrow\left(3+4\right)\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}=3^6\) \(\Rightarrow x+1=6\) \(\Rightarrow x=5\)
Vậy \(x=5\) thì thỏa mãn đề bài.
a)
\(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\\ \Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=2\\\dfrac{x}{2}-\dfrac{1}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{1}{3}+2\\\dfrac{x}{2}=\dfrac{1}{3}-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{7}{3}\\\dfrac{x}{2}=\dfrac{-5}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}.2\\x=\dfrac{-5}{3}.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> x + 2014 = 0
=> x = -2014
vậy x = -2014
c)\(3^{x+2}+4.3^{x+1}=7.3^6\)
\(\Rightarrow3^{x+1}.3+4.3^{x+1}=7.3^6\\ \Rightarrow3^{x+1}\left(3+4\right)=7.3^6\\ \Rightarrow3^{x+1}.7=7.3^6\\ \Rightarrow3^{x+1}=3^6\\ \Rightarrow x+1=6\\ x=6-1\\ x=5\)
vậy x = 5
1. a) \(2009-\left|x-2009\right|=x\)
\(\Rightarrow\left|x-2009\right|=2009-x\)
\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)
\(\Rightarrow x-2009\le0\)
\(\Rightarrow x\le2009\)
Vậy \(x\le2009.\)
b) Ta có: \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}\ge0\forall x\\\left(y-\dfrac{2}{5}\right)^{2008}\ge0\forall y\\\left|x+y-z\right|\ge0\forall x,y,z\end{matrix}\right.\) \(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\forall x,y,z\)
Dấu \("="\) xảy ra khi \(\left[{}\begin{matrix}\left(2x-1\right)^{2008}=0\\\left(y-\dfrac{2}{5}\right)^{2008}=0\\\left|x+y-z\right|=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{5}\\z=\dfrac{9}{10}\end{matrix}\right.\).
Bạn kia làm câu 1 rồi thì mình làm câu 2 nhé!
2. Ta có:\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\)
\(\Rightarrow\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{5b-3c}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{15a-10b}{25}=\dfrac{6c-15a}{9}=\dfrac{15a-10b+6c-15a}{25+9}\)=\(\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)
\(\Rightarrow\dfrac{-5b+3c}{17}=\dfrac{5b-3c}{2}\Rightarrow5b-3c=0\)
=> 5b=3c =>\(\left\{{}\begin{matrix}b=\dfrac{3}{5}c\\a=\dfrac{2}{5}c\end{matrix}\right.\)
=>\(\dfrac{3}{5}c+\dfrac{2}{5}c+c=-50\)
=> \(c\left(\dfrac{3}{5}+\dfrac{2}{5}+1\right)=-50\)
=> 2c = -50
=> c= -25
=>\(\left\{{}\begin{matrix}b=-25.\dfrac{3}{5}=-15\\a=-25.\dfrac{2}{5}=-10\end{matrix}\right.\)
Vậy a= -10; b= -15; c= -25
\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)
\(\Rightarrow\left(\dfrac{x+4}{2009}+1\right)+\left(\dfrac{x+3}{2010}+1\right)=\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+1}{2012}+1\right)\)
\(\Rightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}=\dfrac{x+2013}{2011}+\dfrac{x+2013}{2012}\)
\(\Rightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2011}-\dfrac{x+2013}{2012}=0\)
\(\Rightarrow\left(x+2013\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)
=> x +2013 = 0
=> x = -2013
\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)
\(\Leftrightarrow\dfrac{x+4}{2009}+1+\dfrac{x+3}{2010}+1=\dfrac{x+2}{2011}+1+\dfrac{x+1}{2012}+1\)
\(\Leftrightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}=\dfrac{x+2013}{2011}+\dfrac{x+2013}{2012}\)
\(\Leftrightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2011}-\dfrac{x+2013}{2012}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\right)=0\)
\(\Leftrightarrow x+2013=0\).Do \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)