\(\dfrac{1}{2016}\):2015x=-\(\dfrac{1}{2015}\)

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\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)

\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)

\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)

\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)

Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)

16 tháng 7 2017

help mekhocroi

16 tháng 7 2017

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có:

\(x=\dfrac{2016^{2017}+1}{2016^{2016}+1}< 1\)

\(\Rightarrow x< \dfrac{2016^{2017}+1+2015}{2016^{2016}+1+2015}\Rightarrow x< \dfrac{2016^{2017}+2016}{2016^{2016}+2016}\Rightarrow x< \dfrac{2016\left(2016^{2016}+1\right)}{2016\left(2016^{2015}+1\right)}\Rightarrow x< \dfrac{2016^{2016}+1}{2016^{2015}+1}=y\)

\(\Rightarrow x< y\)

25 tháng 12 2017

\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}=3\)

\(\Rightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)=0\)

\(\Rightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}+\dfrac{x-2017}{2014}=0\)

\(\Rightarrow\left(x-2017\right)\left(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\right)=0\)

\(\dfrac{1}{2016}+\dfrac{1}{2015}+\dfrac{1}{2014}\ne0\) nên \(x-2017=0\Leftrightarrow x=2017\)

25 tháng 12 2017

cảm ơn nhiều

14 tháng 6 2017

a)\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)=\left(x+1\right)\left(\dfrac{1}{13}+\dfrac{1}{14}\right)\)

\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Rightarrow x+1=0\)

\(\Rightarrow x=-1\)

b)\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(1+\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}=1+\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}\)

\(\Rightarrow\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)

Giải tương tự câu a ta được \(x=-2018\)

14 tháng 6 2017

a) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Rightarrow6006\left(x+1\right)+5460\left(x+1\right)+5005\left(x+1\right)=4620\left(x+1\right)+4290\left(x+1\right)\)

\(\Leftrightarrow\left(6006+5460+5005\right)\cdot\left(x+1\right)=\left(4620+4290\right)\cdot\left(x+1\right)\)

\(\Leftrightarrow16471\left(x+1\right)=8910\left(x+1\right)\)

\(\Leftrightarrow16471x+16471=8910x+8910\)

\(\Leftrightarrow16471x-8910x=8910-16471\)

\(\Leftrightarrow7561x=-7561\)

\(\Rightarrow x=-1\)

Vậy \(x=-1\)

b) \(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\Rightarrow4096749040\left(x+4\right)+4094735904\left(x+3\right)=4092704785\left(x+2\right)+4090675680\left(x+1\right)\)

\(\Leftrightarrow4096769040x+16387076160+4094735904x+12284207712=4092704785x+8185409570+4090675680x+4090675680\)

\(\Leftrightarrow8191504944x+28671283872=8183380465x+12276085250\)

\(\Leftrightarrow8191504944x-8183380465x=12276085250-28671283872\)

\(\Leftrightarrow8124479x=-16395198622\)

\(\Rightarrow x=-2018\)

Vậy \(x=-2017\)

P/s: đây không phải cách làm tối ưu, vì vậy mình nghĩ bạn nên tham khảo từ các bài làm khác nhé!

\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)

=>x+2019=0

=>x=-2019

28 tháng 10 2018

ta có :\(\dfrac{y+z-2015x}{x}=\dfrac{z+x-2015y}{y}=\dfrac{z+y-2015z}{z}\)

=>\(\left(\dfrac{y+z-2015}{x}+2016\right)=\left(\dfrac{z+x-2015y}{y}+2016\right)=\left(\dfrac{x+y-2015z}{z}+2016\right)\)

(=)\(\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)

*Nếu x+y+z\(\ne\)0

=>\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)

=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=1.1.1=1

*Nếu x+y+z=0

=>x=y=z

=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=2.2.2=8

12 tháng 11 2018

Thank you!haha

AH
Akai Haruma
Giáo viên
12 tháng 1 2019

Lời giải:

Ta có \(\frac{2016c-2017b}{2015}=\frac{2017a-2015c}{2016}=\frac{2015b-2016a}{2017}\)

\(\Rightarrow \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\( \frac{2015.2016c-2015.2017b}{2015^2}=\frac{2016.2017a-2016.2015c}{2016^2}=\frac{2017.2015b-2017.2016a}{2017^2}\)

\(=\frac{2015.2016c-2015.2017b+2016.2017a-2016.2015c+2017.2015b-2017.2016a}{2015^2+2016^2+2017^2}=0\)

\(\Rightarrow \left\{\begin{matrix} 2015.2016c-2015.2017b=0\\ 2016.2017a-2016.2015c=0\\ 2017.2015b-2016.2016a=0\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 2016c=2017b\\ 2017a=2015c\\ 2015b=2016a\end{matrix}\right.\Rightarrow \frac{a}{2015}=\frac{b}{2016}=\frac{c}{2017}\)

Ta có đpcm.