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c) <=> \(\frac{x+1}{2016}+1+\frac{x+2}{2015}+1\)\(+\frac{x+3}{2014}+1\)= \(\frac{x+4}{2013}+1+\frac{x+5}{2012}+1\)\(+\frac{x+6}{2011}\)
<=> \(\frac{x+1+2016}{2016}+\frac{x+2+2015}{2015}+\frac{x+3+2014}{2014}\) \(=\frac{x+4+2013}{2013}+\frac{x+5+2012}{2012}+\frac{x+6+2011}{2011}\)
<=> \(\frac{x+2017}{2016}+\frac{x+2017}{2015}+\frac{x+2017}{2014}-\frac{x+2017}{2013}-\frac{x+2017}{2012}-\frac{x+2017}{2011}=0\)
<=> \(\left(x+2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
vì \(\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)khác 0
=> \(x+2017=0\) => \(x=-2017\)
Vậy \(S=\left\{-2017\right\}\)
b) \(\frac{x-99}{5}+\frac{x-97}{7}=\frac{x-95}{9}+\frac{x-93}{11}\)
\(\Leftrightarrow\left(\frac{x-99}{5}-1\right)+\left(\frac{x-97}{7}-1\right)=\left(\frac{x-95}{9}-1\right)\)\(+\left(\frac{x-93}{11}-1\right)\)
\(\Leftrightarrow\frac{x-104}{5}+\frac{x-104}{7}-\frac{x-104}{9}-\frac{x-104}{11}=0\)
\(\Leftrightarrow\left(x-104\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)=0\)
Mà \(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\ne0\)
\(\Rightarrow x-104=0\)
\(\Leftrightarrow x=104\)
Vậy ....
a) \(\frac{x+1945}{45}+\frac{x+1954}{54}=\frac{x+1975}{75}+\frac{x+1969}{69}\)
\(\Leftrightarrow\left(\frac{x+1945}{45}-1\right)+\left(\frac{x+1954}{54}-1\right)=\left(\frac{x+1975}{75}-1\right)\)\(+\left(\frac{x+1969}{69}-1\right)\)
\(\Leftrightarrow\frac{x+1900}{45}+\frac{x+1900}{54}-\frac{x+1900}{75}-\frac{x+1900}{69}=0\)
\(\Leftrightarrow\left(x+1900\right)\left(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\right)=0\)
Mà \(\frac{1}{45}+\frac{1}{54}-\frac{1}{75}-\frac{1}{69}\ne0\)
\(\Rightarrow x+1900=0\)
\(\Leftrightarrow x=-1900\)
Vậy ...
(x-1)/2016 +(x-2)/2015 -(x-3)/2014 = (x-4)/2013. =>(x-1)/2016 +(x-2)/2015 = (x-3)/2014 + (x-4)/2013. =>. (X-1)/2016 -1 + (x-2)/2015 -1 = (x -3)/2014 -1 + (x-4)/2013 -1 => (x -2017)/2016 + (x-2017)/2015 -(x-2017)/2014 -(x-2017)/2013 =0. => (x-2017)(1/2016 +1/2015 -1/2014 -1/2013) = 0 => x-2017 =0 => x = 2017
Ta có: \(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\Leftrightarrow\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}-\frac{x-4}{2013}=0\)
\(\Leftrightarrow\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Mà \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)
1) \(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{13}+\frac{x+4}{12}\)
\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{13}-\frac{x+16}{12}=0\)
\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{13}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x=-16\)
2)3)4) tương tự
Gợi ý : 2) cộng 3 vào cả hai vế
3)4) cộng 2 vào cả hai vế
5) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=-3\)
\(\Leftrightarrow\frac{x+21}{20}+\frac{x+21}{19}+\frac{x+21}{18}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}\right)=0\)
\(\Leftrightarrow x=-21\)
6) sửa VT = 4 rồi tương tự câu 5)
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}.\)
\(\left(\frac{x+4}{2012}+1\right)+\left(\frac{x+3}{2013}+1\right)=\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)\)
\(\left(\frac{x+4}{2012}+\frac{2012}{2012}\right)+\left(\frac{x+3}{2013}+\frac{2013}{2013}\right)=\left(\frac{x+2}{2014}+\frac{2014}{2014}\right)+\left(\frac{x+1}{2015}+\frac{2015}{2015}\right)\)
\(\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
\(\frac{x+2016}{2012}+\frac{x+2016}{2013}-\frac{x+2016}{2014}-\frac{x+2016}{2015}=0\)
\(\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
\(\Rightarrow x+2016=0\Rightarrow x=\left(-2016\right)\)
\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)
\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)
Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức
\(a)\) \(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)
\(\Leftrightarrow\)\(\frac{x-1+2011}{2011}+\frac{x-2+2012}{2012}=\frac{x-3+2013}{2013}+\frac{x-4+2014}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}-\frac{x+2010}{2013}-\frac{x+2010}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow\)\(x=2010\)
Vậy \(x=2010\)
Chúc bạn học tốt ~
Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.
Cộng vào mỗi phân số thêm 1 đơn vị được:
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).
Tới đây tự làm tiếp nhá.
Ta có :
\(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\Leftrightarrow\)\(\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+4+2014}{2014}+\frac{x+3+2015}{2015}=\frac{x+8+2010}{2010}+\frac{x+7+2011}{2011}\)
\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Leftrightarrow\)\(\frac{x+2018}{2014}+\frac{x+2018}{2015}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
Nên \(x-2018=0\)
\(\Leftrightarrow\)\(x=2018\)
Vậy \(x=2018\)
Chúc bạn học tốt ~
Ta có: \(\frac{x+4}{2014}+\frac{x+3}{2015}=\frac{x+7}{2011}+\frac{x+8}{2010}\)
\(\Rightarrow\left(\frac{x+4}{2014}+1\right)+\left(\frac{x+3}{2015}+1\right)=\left(\frac{x+7}{2011}+1\right)+\left(\frac{x+8}{2010}+1\right)\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}=\frac{x+2018}{2011}+\frac{x+2018}{2010}\)
\(\Rightarrow\frac{x+2018}{2014}+\frac{x+2018}{2013}-\frac{x+2018}{2011}-\frac{x+2018}{2010}=0\)
\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2018=0\Rightarrow x=-2018\)
Chúc bn hc tốt! ^_^