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Phong Nguyễn Trần
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Trần Khởi My
11 tháng 4 2017 lúc 12:57

hình như phải là dấu = bn ak

nguyễn Thị Bích Ngọc
11 tháng 4 2017 lúc 16:48

x=y=0

nguyễn Thị Bích Ngọc
11 tháng 4 2017 lúc 16:49

mà z đâu ra thế

Bla bla bla
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Trần Tuấn Hoàng
14 tháng 12 2023 lúc 19:24

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

Nàng tiên cá
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♥ℒℴѵe♥
20 tháng 10 2017 lúc 20:12

\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}=\frac{x-4}{2011}\)

\(\frac{x-1}{2014}+\frac{x-2}{2013}-\frac{x-3}{2012}-\frac{x-4}{2011}=0\)

\(\left(\frac{x-1}{2014}-1\right)+\left(\frac{x-2}{2013}-1\right)-\left(\frac{x-3}{2012}-1\right)-\left(\frac{x-4}{2011}-1\right)=0\)

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

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

Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\)

\(\Rightarrow x-2015=0\)

\(x=0+2015\)

\(x=2015\)

Ngốc Trần
20 tháng 10 2017 lúc 20:19

\(x=2015\)

chi phạm
11 tháng 2 2020 lúc 12:53

x=2015 nha

Khách vãng lai đã xóa
Nữ hoàng sến súa là ta
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Nguyệt
13 tháng 12 2018 lúc 22:28

\(\Rightarrow\frac{x}{2010}+\frac{x+1}{2011}+\frac{x+2}{2012}+\frac{x+3}{2013}+\frac{x+4}{2014}-5=0\)

\(\left(\frac{x}{2010}-1\right)+\left(\frac{x+1}{2011}-1\right)+\left(\frac{x+2}{2012}-1\right)\)\(+\left(\frac{x+3}{2013}-1\right)+\left(\frac{x+4}{2014}-1\right)=0\)

\(\frac{x-2010}{2010}+\frac{x-2010}{2011}+\frac{x-2010}{2012}+\frac{x-2010}{2013}+\frac{x-2010}{2014}=0\)

\(\left(x-2010\right).\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)=0\)

mà \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)

Vậy x=-2010

oOo NhỎ tHiêN cHỉ HạC oO...
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Phương Trâm
11 tháng 8 2017 lúc 21:34

\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)

\(\Leftrightarrow\dfrac{x+4}{2012}+1+\dfrac{x+3}{2013}+1=\dfrac{x+2}{2014}+1+\dfrac{x+1}{2015}\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\left(\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\right)=0\)

\(\Leftrightarrow x+2016.\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\right)\)

\(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}\ne0\)

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

Vậy \(x=-2016\) tại biểu thức \(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)

Sakura Nguyen
11 tháng 8 2017 lúc 21:36

Theo đề ta có: x+4/2012+x+3/2013=x+2/2014+x+1/2015
=>x+4/2012+x+3/2013-x+2/2014+x+1/2015=0
=>( x+4/2012+1)+(x+3/2013+1)-(x+2/2014+1)+(x+1/2015+1)
=>x+2016/2012+x+2016/2013-x+2016/2014-x+2016/2015=0
=>x+2016.(1/2012+1/2013-1/2014-1/2015)=0
Do 1/2012+1/2013-1/2014-1/2015>0
nên x+2016=0
=>x=-2016
Vậy x=-2016

Nguyễn Thanh Hằng
11 tháng 8 2017 lúc 21:36

\(\dfrac{x+4}{2012}+\dfrac{x+3}{2013}=\dfrac{x+2}{2014}+\dfrac{x+1}{2015}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2012}+1\right)+\left(\dfrac{x+3}{2013}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2014}+1\right)\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}=\dfrac{x+2016}{2014}+\dfrac{x+2016}{2015}\)

\(\Leftrightarrow\dfrac{x+2016}{2012}+\dfrac{x+2016}{2013}-\dfrac{x+2016}{2014}-\dfrac{x+2016}{2015}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)

\(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\ne0\Leftrightarrow x+2016=0\)

\(\Leftrightarrow x=-2016\)

⊰⊹🅼🅸🅸🅽🅷☠☆
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.Gió..
11 tháng 5 2023 lúc 20:15

\(\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}=5\)

\(\Leftrightarrow\dfrac{x}{2012}+\dfrac{x+1}{2013}+\dfrac{x+2}{2014}+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-5=0\)

\(\Leftrightarrow\dfrac{x}{2012}-1+\dfrac{x+1}{2013}-1+\dfrac{x+2}{2014}-1+\dfrac{x+3}{2015}+\dfrac{x+4}{2016}-1=0\)

\(\Leftrightarrow\dfrac{x-2012}{2012}+\dfrac{x-2012}{2013}+\dfrac{x-2012}{2014}+\dfrac{x-2012}{2015}+\dfrac{x-2012}{2016}=0\)

\(\Leftrightarrow\left(x-12\right).\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}\right)=0\)

\(\Leftrightarrow x-12=0\)

\(\Leftrightarrow x=12\)

___Vương Tuấn Khải___
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ngo thi phuong
11 tháng 10 2017 lúc 13:00

Hỏi đáp Toán

Nhã Doanh
Xem chi tiết
Giang
20 tháng 9 2017 lúc 20:51

Giải:

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)

\(\Leftrightarrow2+\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=2+\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)

\(\Leftrightarrow1+\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}=1+\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}\)

\(\Leftrightarrow\left(1+\dfrac{x+1}{2015}\right)+\left(1+\dfrac{x+2}{2014}\right)=\left(1+\dfrac{x+3}{2013}\right)+\left(1+\dfrac{x+4}{2012}\right)\)

\(\Leftrightarrow\dfrac{x+1+2015}{2015}+\dfrac{x+2+2014}{2014}=\dfrac{x+3+2013}{2013}+\dfrac{x+4+2012}{2012}\)

\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)

\(\Leftrightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)

\(\Leftrightarrow\left(x+2016\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)

\(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)

Nên \(x+2016=0\)

\(\Leftrightarrow x=0-2016\)

\(\Leftrightarrow x=-2016\)

Vậy ...

Chúc bạn học tốt!

TM Vô Danh
20 tháng 9 2017 lúc 21:10

\(\dfrac{x+1}{2015}+\dfrac{x+2}{2014}=\dfrac{x+3}{2013}+\dfrac{x+4}{2012}\)

\(\Rightarrow\dfrac{x+1}{2015}+1+\dfrac{x+2}{2014}+1=\dfrac{x+3}{2013}+1+\dfrac{x+4}{2012}+1\)

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}=\dfrac{x+2016}{2013}+\dfrac{x+2016}{2012}\)

\(\Rightarrow\dfrac{x+2016}{2015}+\dfrac{x+2016}{2014}-\dfrac{x+2016}{2013}-\dfrac{x+2016}{2012}=0\)

\(\Rightarrow\left(x+2016\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\right)=0\)

do \(\dfrac{1}{2015}+\dfrac{1}{2014}-\dfrac{1}{2013}-\dfrac{1}{2012}\ne0\)

\(\Rightarrow x+2016=0\Rightarrow x=2016\)

váy x=2016

 Mashiro Shiina
26 tháng 4 2018 lúc 10:15

Bây h mới biết hồi đó you ngok v:3

Lê Vũ Anh Thư
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Shizadon
16 tháng 11 2017 lúc 21:45

Ta có : \(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}=4\)

\(\dfrac{x-3}{2015}+\dfrac{x-4}{2014}+\dfrac{x-5}{2013}+\dfrac{x-6}{2012}-4=0\)

\(\dfrac{x-3}{2015}-1+\dfrac{x-4}{2014}-1+\dfrac{x-5}{2013}-1+\dfrac{x-6}{2012}-1=0\)

\(\dfrac{x-2018}{2015}+\dfrac{x-2018}{2014}+\dfrac{x-2018}{2013}+\dfrac{x-2018}{2012}=0\)

\(\left(x-2018\right).\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}\right)=0\)

\(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}>0\)

=> x - 2018 = 0

x = 0 + 2018

x = 2018

Vậy x = 2018