NN
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
27 tháng 9 2020
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2011\\y\ge2012\\z\ge2013\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-2011}\ge0\\b=\sqrt{y-2012}\ge0\\c=\sqrt{z-2013}\ge0\end{matrix}\right.\) ta có :
\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)
\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)
19 tháng 9 2019
= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)
= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)
= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)
= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)
= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)
NN
2
15 tháng 8 2017
\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)
\(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)=\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)
\(\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+2015=0\\\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}=0\end{cases}}\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\Rightarrow x+2015=0\Rightarrow x=-2015\)
Vậy x = 2015 nha bn
TN
15 tháng 8 2017
\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)
\(\Rightarrow\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)=\left(\frac{x+3}{2012}+1\right)+\left(\frac{x+4}{2011}+1\right)\)
\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{x+2011}\)
\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)
\(\Rightarrow\left(x+2015\right).\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\Rightarrow\left(x-2015\right)=0\)
\(\Rightarrow x=0+2015\) =2015
Đúng thì k ủng hộ mik nha mn!
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 \(a\ge0\Rightarrow a=1\)
\(\Rightarrow x^2+2x+1=1\)
\(x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2,0\right\}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
16 tháng 2 2023
A = 12 - 22 + 32 - 42 + 52 - 62 + 72 - .......- 582 + 592
A = 12 + ( 32 - 22) + ( 52 - 42) + (72 - 62) +....+ ( 592 - 582)
A = 1 + ( 3-2)(2+3) + (5-4)(4+5) + (7-6)(6+7)+....+(59-58)(58+59)
A = 1 + 2 + 3 + 4 + 5 + 6 + 7 + ....+ 58 + 59
A = ( 59 + 1).{ (59 - 1): 1 + 1 } : 2
A = 1770
B = \(\dfrac{2^{2016}-2^{2015}+2^{2014}-2^{2013}+2^{2012}-2^{2011}+2^{2010}-2^{2009}}{2^{2008}}\)
Đặt tử số là A
ta có
A = 22016 - 22015+22014 - 22013 + 22012 - 22011 + 22010- 22009
2 A= 22017- 22016 + 22015- 22014 +22013-22012 + 22011 - 22010
2A + A = 22017 - 22009
3A = 22017 - 22009
A = (22017 - 22009):3
B = A : 8 = (22017- 22009) : 3 : 8
B = (22017 - 22009) : 24
\(P=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{2011.2012.2013}\)
\(2P=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{2011.2012.2013}\)
\(=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2011.2012}-\frac{1}{2012.2013}\right)\)
\(=\frac{1}{1.2}-\frac{1}{2012.2013}=\frac{2013.2012-2}{2025078}=\frac{4050154}{2025078}\)
\(\Rightarrow P=\frac{4050154}{2025078}:2\)
oh mình tính sai :
\(2P=\frac{1}{1.2}-\frac{1}{2013.2012}=\frac{2013.2012-2025078}{2013.2012}=\frac{4050156-2025078}{4050156}=\frac{2025078}{4050156}\)
\(\Rightarrow P=\frac{2025078}{4050156}:2=\frac{1}{4}\)