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AN
1
HP
25 tháng 9 2018
Vì \(x^{2015}+y^{2015}=x^{2016}+y^{2016}=x^{2017}+y^{2017}\)
\(\Rightarrow x=y=1\) hoặc \(x=y=0\)
Với \(x=y=1\)
\(S=2018\left(1^{2018}+1^{2018}\right)\)
\(S=2018.2\)
\(S=4036\)
Với \(x=y=0\)
\(S=2018\left(0^{2018}+0^{2018}\right)\)
\(S=0\)
KC
3
8 tháng 3 2018
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
<=>\(\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{2017}-1+\dfrac{x+4}{2018}-1\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
vì 1/2015+1/2016-1/2017-1/2018 khác 0
=>x-2014=0<=>x=2014
vậy.....................
chúc bạn học totts ^^
ND
8 tháng 3 2018
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
\(\Leftrightarrow\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{x017}-1+\dfrac{x+4}{2018}-1\)
\(\Leftrightarrow\dfrac{x+1-2015}{2015}+\dfrac{x+2-2016}{2016}=\dfrac{x+3-2017}{2017}+\dfrac{x+4-2018}{2018}\)\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Vì: \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy........
DS
0
TN
16 tháng 6 2016
a)ko bít đề bắt làm j
b)Px=x(1+x+x2+...+x2015+x2018)
Px=x+x2+...+x2017
Px-P=(x+x2+...+x2017)-(1+x+x2+...+x2015+x2018)
P(x-1)=x2017-1
P=(x2017-1)/(x-1)
16 tháng 7 2017
Ta có:
\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)
\(\Leftrightarrow\left(\dfrac{x}{2014}-1\right)+\left(\dfrac{x+1}{2015}-1\right)+\left(\dfrac{x+2}{2016}-1\right)+\left(\dfrac{x+3}{2017}-1\right)+\left(\dfrac{x+4}{2018}-1\right)=0\)\(\Leftrightarrow\dfrac{x-2014}{2014}+\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}+\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}=0\)\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)=0\) (1)
Mà \(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}>0\) (2)
Từ (1) và (2) => \(x-2014=0\) \(\Leftrightarrow x=2014\)
NT
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 3 2019
\(a;b;c\ne0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2018}=\frac{1}{a+b+c}\)\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\Leftrightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b=0\\ab=-c\left(a+b+c\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\ab+ac+bc+c^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\\left(a+c\right)\left(b+c\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\a+c=0\\b+c=0\end{matrix}\right.\)
\(M=\left(a^{2015}+b^{2015}\right)\left(a^{2017}+b^{2017}\right)\left(a^{2019}+b^{2019}\right)\)
- Nếu \(a+b=0\Rightarrow M=0\)
- Nếu \(\left[{}\begin{matrix}a+c=0\\b+c=0\end{matrix}\right.\) thì ko tính được giá trị cụ thể của M
Khi đó \(\left[{}\begin{matrix}M=\left(2018^{2015}+b^{2015}\right)\left(2018^{2017}+b^{2017}\right)\left(2018^{2019}+b^{2019}\right)\\M=\left(2018^{2015}+a^{2015}\right)\left(2018^{2017}+a^{2017}\right)\left(2018^{2019}+a^{2019}\right)\end{matrix}\right.\)
OM
5 tháng 4 2020
a, Làm
\(\frac{x+1}{2020}+\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x+5}{2016}+\frac{x+6}{2015}\)
<=>\(\frac{x+2021}{2020}+\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2016}+\frac{x+2021}{2015}\)
<=>\(\left(x+2021\right)\left(\frac{1}{2020}+\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
<=> x+2021=0
<=> x=-2021
Kl:......................
b, Làmmmmm
\(\frac{2-x}{2004}-1=\frac{1-x}{2005}-\frac{x}{2006}\)
<=> \(\frac{2006-x}{2004}=\frac{2006-x}{2005}+\frac{2006-x}{2006}\)
<=> \(\left(2006-x\right)\left(\frac{1}{2004}-\frac{1}{2005}-\frac{1}{2006}\right)=0< =>2006-x=0\)
<=> x=2006
Kl:..............
UK
19 tháng 8 2017
b) \(x,y\ge1\Rightarrow xy\ge1\)
BĐT đã cho tương đương với:
\(\left(\dfrac{1}{1+x^2}-\dfrac{1}{1+xy}\right)+\left(\dfrac{1}{1+y^2}-\dfrac{1}{1+xy}\right)\ge0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{xy-y^2}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow+\dfrac{x\left(y-x\right)}{\left(1+x^2\right)\left(1+xy\right)}+\dfrac{y\left(x-y\right)}{\left(1+y^2\right)\left(1+xy\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(y-x\right)^2\left(xy-1\right)}{\left(1+x^2\right)\left(1+y^2\right)\left(1+xy\right)}\ge0\)
BĐT cuối luôn đúng nên ta có đpcm
Đẳng thức xảy ra khi x=y hoặc xy=1
