Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: 12-22+32-............+20152
C=20152-20142+...............+32-22+12
C=(2015+2014)(2015-2014)+(2013+2012)(2013-2012)+...........+(3+2)(3-2)+12
C=2015+2014+2013+.........+3+2+12=2015+2014+2013+............+1
C=2016.2015:2
C=1008.2015
C=??????? bạn tự dùng máy tính
\(C=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2013-2014\right)\left(2013+2014\right)+2015^2\)
\(=2015^2-\left(1+2+3+4+...+2013+2014\right)\)
\(=2015^2-\dfrac{2015\cdot2014}{2}=2031120\)
Số số hạng của tổng B là:
\(\frac{\left(2015-1\right)}{1}+1=2015\)(số hạng)
\(B=\frac{\left(1+2015\right)\cdot2015}{2}=2031120\)
\(A=\left(1^2-2^2\right)+\left(3^2-4^2\right)+\left(5^2-6^2\right)+...+\left(2013^2-2014^2\right)+2015^2\)
\(A=\left(-3\right)+\left(-7\right)+\left(-11\right)+...+\left(-4027\right)+4060225\)
Số số hạng của tổng A thuộc nguyên âm là:
\(\frac{2014}{2}=1007\)(số hạng)
\(A=\frac{\left(-3\right)+\left(-4027\right)\cdot1007}{2}+4060225\)
\(A=\left(-2029105\right)+4060225\)
\(A=2031120\)
Mà \(2031120=2031120\)
\(\Rightarrow A=B\)
\(A=1^2-2^2+3^2-4^2+...-2014^2+2015^2\)
\(A=1+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2015^2-2014^2\right)\)
\(A=1+\left(3-2\right).\left(2+3\right)+\left(4-5\right).\left(4+5\right)+...+\left(2015-2014\right).\left(2014+2015\right)\)
\(A=1+2+3+4+...+2015=B\)
(Mình giải theo cách lớp 8 nhé)
\(A=1^2-2^2+3^2-4^2+...+2015^2\)
\(=1+\left(3^2-2^2\right)+\left(5^2-4^2\right)+...+\left(2015^2-2014^2\right)\)
\(=1+\left(3-2\right)\left(3+2\right)+\left(5-4\right)\left(5+4\right)+...+\left(2015-2014\right)\left(2015+2014\right)\)
\(=1+\left(2+3\right)+\left(4+5\right)+...+\left(2014+2015\right)\)
\(=1+2+3+...+2015=B\)
\(\Leftrightarrow A=B\)
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 ~
Ta có: C=12-22+32-42+...+20152
=(20152-20142)+(20132-20122)+...+(32-22)+12
=(2015+2014)(2015-2014)+(2013+2012)(2013-2012)+...+(3+2)(3-2)+1
=(2015+2014).1+(2013+2012).1+...+(3+2).1+1
=1+2+3+...+2012+2013+2014+2015
=(2015+1)[(2015-1)/1+1]/2
=2031120
C=1^2-2^2+3^2-4^2+...+2013^2-2014^2+2015^2
=(2015^2-2014^2)+(2013^2-2012^2)+...+(5^2-4^2)+(3^2-2^2)+1^2
=(2015-2014)(2014+2015)+(2013-2012)(2013+2012)+..+(5-4)(5+4)+(3-2)(3+2)+1
=4029+4025+...+9+5+1
số số hạng (4029-1):4+1=1008
tổng là [(4029+1).1008]:2=2031120
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
b) \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}=\dfrac{1}{18}\\< =>\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ < =>\dfrac{1}{x+1}-\dfrac{1}{x+5}=\dfrac{1}{18}\\ quyđồngmẫuvàkhửmẫu\\ x^{2^{ }}+6x-27=0\\ giảipttìmđược:x=3;x=-9\)
a) \(\frac{x-2015}{1}+\frac{x-2014}{2}+\frac{x-2013}{3}+...+\frac{x-1}{2015}+\frac{x}{2016}=0\\ \Leftrightarrow\frac{x-2015}{1}-1+\frac{x-2014}{2}-1+...+\frac{x-1}{2015}-1+\frac{x}{2016}-1=-2016\)
\(\Leftrightarrow\frac{\left(x-2016\right).1}{1}+\frac{\left(x-2016\right).1}{2}+\frac{\left(x-2016\right).1}{3}+...+\frac{\left(x-2016\right).1}{2015}+\frac{\left(x-2016\right).1}{2016}=-2016\)
\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)=-2016\)
tới đây mình chịu. mình nghĩ là phương trình bạn cho là bằng 2016 chứ, như thế giải mới được, còn như này thì mình bó tay
b)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{8}\\ \Leftrightarrow\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{8}\\ \Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{4}{\left(x+2\right)\left(x+6\right)}=\frac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\)
\(\Leftrightarrow x^2+8x+12-32=0\\ \Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-10\\x=2\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={-10;2}
\(C=\left(1^2-2^2\right)+\left(3^2-4^2\right)+....+\left(2013^2-2014^2\right)+2015^2\)
\(C=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+....+\left(2013-2014\right)\left(2013+2014\right)+2015^2\)
\(C=-\left(1+2\right)-\left(3+4\right)-....-\left(2013+2014\right)+2015^2\)
\(C=-\left(1+2+3+4+...+2014\right)+2015^2\)
\(C=-\dfrac{\left(2014+1\right)2014}{2}+2015^2\)
\(C=-2015.1007+2015^2\)
\(C=2015\left(2015-1007\right)=2015.1008\)