Cho \(\frac{2010\cdot c-2011\cdot b}{2009}=\frac{2011\cdot a-2009\cdot c}{2010}=\frac{2009\cdot b-2010\cdot c}{2011}\)
C/m \(\frac{a}{2009}=\frac{b}{2010}=\frac{c}{2011}\)
K
Khách
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17 tháng 10 2018
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
N
26 tháng 2 2019
\(\frac{b-2011}{c-2010}:\frac{2011-b}{2010-c}=\frac{b-2011}{c-2010}\cdot\frac{-\left(c-2010\right)}{-\left(b-2011\right)}=1\)
\(\frac{a-2009}{b-2011}=\frac{2010-c}{2009-a}=\frac{-\left(c-2010\right)}{-\left(a-2009\right)}=\frac{c-2010}{a-2009}=1\Rightarrow a-2009=c-2010=b-2011\)
\(\Rightarrow a=c-1=b-2\Rightarrow c=b-1\Rightarrow\frac{b}{c}=\frac{b}{b-1}\)=.=' ko chắc lăm
8 tháng 5 2020
\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\) ( có lẽ đề như này )
\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)
\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)
\(\Leftrightarrow x=2014\)
...
Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)
\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow3.18=x^2+4x+7x+28\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
MT
14 tháng 1 2016
\(2009-\frac{2010}{3}-\frac{2010}{6}-\frac{2010}{15}-...-\frac{2010}{45}\)
\(=2009-2010.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{15}+...+\frac{1}{45}\right)\)
\(=2009-2010.\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{90}\right)\)
Có vấn đề chỗ 2010/15 bạn xem lại
T
20 tháng 6 2019
#)Giải :
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)
\(=0\)
20 tháng 6 2019
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)
=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)
\(=0\)
2 tháng 9 2022
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
