a) (x-5)^x+2015 - (x-5)^x+2014  =0

  (x-5)^x+2014(x-5 -1) =0

+ x -5 =0 => x =5

+ x -6 =0 => x =6

Vậy x = 5 hoặc x =6

thêm số hạng 1 vào bên phải nha

Xét \( A = 1 + \dfrac{{2014}}{2} + \dfrac{{2015}}{3} + ... + \dfrac{{4023}}{{2011}} + \dfrac{{4024}}{{2012}}\\ \)

\(\Rightarrow A - 2012 = \left( {\dfrac{{2014}}{2} - 1} \right) + \left( {\dfrac{{2015}}{3} - 1} \right) + ... + \left( {\dfrac{{4024}}{{2012}} - 1} \right)\\ \Rightarrow A - 2012 = \dfrac{{2012}}{2} + \dfrac{{2012}}{3} + ... + \dfrac{{2012}}{{2012}}\\ \Rightarrow A - 2012 = 2012\left( {\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow A = 2012\left( {1 + \dfrac{1}{2} + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow \left( {1 + \dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{{2012}}} \right)503x = 2012\left( {1 + ... + \dfrac{1}{{2012}}} \right)\\ \Rightarrow x = \dfrac{{2012}}{{503}} = 4 \)

\(1\frac{1}{5}\)X \(1\frac{1}{6}\) X \(1\frac{1}{7}\) X \(...\)X \(1\frac{1}{2014}\) = x

x = \(\frac{6}{5}\) X \(\frac{7}{6}\) X \(\frac{8}{7}\) X \(...\)X \(\frac{2015}{2014}\)

x = \(\frac{6\cdot7\cdot8\cdot...\cdot2015}{5\cdot6\cdot7\cdot...2014}\)

x = \(\frac{2015}{5}\)

x = 403

6/5 × 7/6 × 8/7 × ... × 2015/2014 = x

2015/5 = x

x = 403

Vậy x = 403

Ủng hộ mk nha ♡_♥☆_★^_-

\(\Rightarrow x=\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times.......\times\frac{2015}{2014}\)

\(\Rightarrow x=\frac{6\times7\times8\times......\times2015}{5\times6\times7\times......\times2014}\)

\(\Rightarrow x=\frac{2015}{5}=403\)

\(\frac{x+5}{2015}+\frac{x+6}{2014}+\frac{x+7}{2013}+\frac{x+8}{2012}+\frac{x+9}{2011}+5=0\)

\(\Rightarrow1+\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1+\frac{x+8}{2012}+1+\frac{x+9}{2011}=0\)

\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}+\frac{x+2020}{2012}+\frac{x+2020}{2011}=0\)

\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)

\(\Rightarrow x+2020=0\)

\(\Rightarrow x=-2020\)

Study well 

chuyên toán thcs

Thiếu giải thích

Trả lời 

không cần giải thích nha 

\(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}=0\right)\)

\(\Rightarrow\left(x+2020\right)=0:\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)

\(\Rightarrow\left(x+2020\right)=0\)

Tôi ngại chuyển vế nên làm tắt có cần giả thích cái j đâu

helpppppppppppppppp

Bài làm:

Pt <=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-5}{2016}-1\right)+\left(\frac{x-7}{2014}-1\right)=4-4\)

\(\Leftrightarrow\frac{x-2021}{2020}+\frac{x-2021}{2018}+\frac{x-2021}{2016}+\frac{x-2021}{2014}=0\)

\(\Rightarrow x-2021=0\Rightarrow x=2021\)

Ta có :\(\frac{x-1}{2020}+\frac{x-3}{2018}+\frac{x-5}{2016}+\frac{x-7}{2014}=4\)

=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-5}{2016}-1\right)+\left(\frac{x-7}{2014}-1\right)=0\)

=> \(\frac{x-2021}{2020}+\frac{x-2021}{2018}+\frac{x-2021}{2016}+\frac{x-2021}{2014}=0\)

=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2018}+\frac{1}{2016}+\frac{1}{2014}\right)=0\)

Vì \(\frac{1}{2020}+\frac{1}{2018}+\frac{1}{2016}+\frac{1}{2014}\ne0\)

=> x - 2021 = 0

=> x = 2021

Vậy x = 2021

\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)

\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)

\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)

bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.

Bài 1 : 

Ta có  : 

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

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

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

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

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

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

\(\Rightarrow x+4024=0\)

\(\Rightarrow x=-4024\)

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 \(a\ge0\Rightarrow a=1\)

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

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

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

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