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12 tháng 2 2019

\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)

\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0

\(\Leftrightarrow x=2013\)

Vậy ........

7 tháng 5 2018

Ta có : 

\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=2012\)

\(\Leftrightarrow\)\(\frac{x-1-2012}{2012}+\frac{x-2-2011}{2011}+\frac{x-3-2010}{2010}+...+\frac{x-2012-1}{1}=0\)

\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\ne0\)

Nên \(x-2013=0\)

\(\Leftrightarrow\)\(x=2013\)

Vậy \(x=2013\)

Chúc bạn học tốt ~ 

7 tháng 5 2018

\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1+2012=2012\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

\(\Leftrightarrow x=2013\)

7 tháng 1 2018

Phương trình đã cho tương đương với :

\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2012=2012\)

\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

Tìm x theo như toán lớp 6 nha

\(x-2013=0\)

\(\Leftrightarrow\)\(x=2013\)

7 tháng 1 2018

ta có pt 

<=>\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1=0\)

<=>\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

<=>\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\Leftrightarrow x-2013=0\Leftrightarrow x=2013\)

^_^

22 tháng 4 2020

Bài 1 : 

Ta có  : 

\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)

\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)

\(+\left(\frac{x+2013}{2011}+1\right)\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)

\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)

\(\Rightarrow x+4024=0\)

\(\Rightarrow x=-4024\)

22 tháng 4 2020

Bài 2 : 

Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)

=> Phương trình trở thành 

\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)

\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)

\(\Rightarrow5a^2+3a-8=0\)

\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)

Vì \(a\ge0\Rightarrow a=1\)

\(\Rightarrow x^2+2x+1=1\)

\(x^2+2x=0\)

\(\Rightarrow x\left(x+2\right)=0\)

\(\Rightarrow x\in\left\{-2,0\right\}\)

11 tháng 7 2016

a) \(=-7\left(x^2-\frac{10}{7}x+\frac{2016}{7}\right)\)

      \(=-7\left(x^2-2.\frac{5}{7}x+\frac{25}{49}+\frac{14087}{49}\right)\)

       \(=-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\)

ta có

\(\left(x-\frac{5}{7}\right)^2\ge0\)với mọi x

\(=>-7\left(x-\frac{5}{7}\right)^2\le0\)(nhân cả hai vế với -7)

\(=>-7\left(x-\frac{5}{7}\right)^2-\frac{14087}{7}\le-\frac{14087}{7}\)

trường hợp dấu "=" xảy ra khi và chỉ khi

\(\left(x-\frac{5}{7}\right)^2=0\)

\(=>x-\frac{5}{7}=0\)

\(=>x=\frac{5}{7}\)

vậy GTLN cảu biểu thức là \(-\frac{14087}{7}\) khi và chỉ khi x= \(\frac{5}{7}\)

4 tháng 2 2018

Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+...+\frac{x-2012}{1}=2012\)

\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+....+\frac{x-2012}{1}-1=2012-2012\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2011}+....+1\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\ne0\)

\(\Rightarrow x+2013=0\)

\(\Rightarrow x=2013\)

Vậy x = 2013

4 tháng 2 2018

PT đã cho tương đương với:

      \(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2010=2012\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

\(\Leftrightarrow x=2013\)

5 tháng 2 2018

Ta có phương trình : 

\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+....+\frac{x-2012}{1}=2012\)

Ta thấy phương trình đã cho tương ứng với phương trình : 

\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)+2012=2012\)

\(\Rightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Rightarrow\left(x-2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+....+1\right)=0\)

Mặt khác \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\ne0\)

Do đó \(\Rightarrow x-2013=0\Rightarrow x=2013\)

Do vậy \(x=2013\)thoả mãn phương trình ban đầu 

5 tháng 2 2018

\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2000}+.....+\frac{x-2012}{1}=2012\)

\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+........+\frac{x-2012}{1}-2012=0\)

\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+......+\left(\frac{x-2012}{1}-1\right)=0\)

\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+......+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+.....+1\right)=0\)

Mà \(\frac{1}{2012}+\frac{1}{2011}+....+1\ne0\)

Vậy ...

\(\Leftrightarrow x=2013\)

\(\Leftrightarrow x-2013=0\)

14 tháng 3 2019

\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)

\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

tự làm nốt~

14 tháng 3 2019

kudo shinichi làm sai ở chỗ:

\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé

7 tháng 3 2017

\(\frac{x-2}{2012}+\frac{x-3}{2011}+\frac{x-4}{2010}+\frac{x-2029}{5}=0\)

\(\Leftrightarrow\frac{x-2}{2012}-1+\frac{x-3}{2011}-1+\frac{x-4}{2010}-1+\frac{x-2029}{5}+3=0\)

\(\Leftrightarrow\frac{x-2014}{2012}+\frac{x-2014}{2011}+\frac{x-2014}{2010}+\frac{x-2014}{5}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow x-2014=0\).Do \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\ne0\)

\(\Leftrightarrow x=2014\)

27 tháng 2 2019

Ta có:\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)

\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)

\(\Rightarrow\left(x-2014\right).\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)\)

\(\Rightarrow x-2014=0\)( vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\))

\(\Rightarrow x=2014\)

Vậy x= 2014.