Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
C = 12 - 22 + 32 - 42 + 52 - 62 + ... + 20132 - 20142 + 20152
C = (1 - 2).(1 + 2) + (3 - 4).(3 + 4) + (5 - 6).(5 + 6) + ... + (2013 - 2014).(2013 + 2014) + 20152
C = -(1 + 2) + [-(3 + 4)] + [-(5 + 6)] + ... + [-(2013 + 2014)] + 4060225
C = -(1 + 2 + 3 + 4 + 5 + 6 + ... + 2013 + 2014) + 4060225
C = -(1 + 2014).2014:2 + 4060225
C = -2015.1007 + 4060225
C = -2029105 + 4060225
C = 2031120
\(C=1-2^2+3^2-4^2+...+2013^2-2014^2+2015^2\)
\(\Leftrightarrow C=2015^2+\left(1-2014^2\right)-\left(2^2-2013^2\right)+\left(3^2-2012^2\right)-...\)
\(\Leftrightarrow C=2015^2+\left(1+2014\right)\left(1-2014\right)-\left(2+2013\right)\left(2-2013\right)+\left(3+2012\right)\left(3-1012\right)-...\)\(\Leftrightarrow C=2015^2+\left[2015.\left(-2013\right)\right]-\left[2015.\left(-2013\right)\right]+...\)
\(\Leftrightarrow C=2015^2\)
(?)
C=(1-2)(1+2)+(3-4)(3+4)+...+(2013-2014)(2013+2014)+2015^2
=2015^2-(1+2+3+...+2013+2014)
=2015^2-2014*2013/2
=2033134
Bài 2:
a:
Sửa đề: B=(3x+5)^2+(3x-5)^2-2(3x+5)(3x-5)
=(3x+5-3x+5)^2
=10^2
=100
b: =(1-2)(1+2)+(3-4)(3+4)+...+(2013-2014)(2013+2014)+2015^2
=2015^2-(1+2+...+2013+2014)
=2031120
Bài 3 :
\(\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-3}{2014}+\frac{x-4}{2013}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)=\left(\frac{x-3}{2014}-1\right)+\left(\frac{x-4}{2013}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1-2016}{2016}+\frac{x-2-2015}{2015}=\frac{x-3-2014}{2014}+\frac{x-4-2013}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2014}+\frac{x-2017}{2013}\)
\(\Leftrightarrow\)\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\Leftrightarrow\)\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\)
Nên \(x-2017=0\)
\(\Rightarrow\)\(x=2017\)
Vậy \(x=2017\)
Chúc bạn học tốt ~
Bài 1 :
\(\left(8x-5\right)\left(x^2+2014\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x-5=0\\x^2+2014=0\end{cases}\Leftrightarrow\orbr{\begin{cases}8x=0+5\\x^2=0-2014\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}8x=5\\x^2=-2014\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\sqrt{-2014}\left(loai\right)\end{cases}}}\)
Vậy \(x=\frac{5}{8}\)
Chúc bạn học tốt ~