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A=\(\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\cdot\cdot\dfrac{-2015}{2016}\)
=\(-\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\cdot\cdot\dfrac{2015}{2016}\)
=\(\dfrac{-1}{2016}>\dfrac{-1}{2015}\)
Vậy\(A>\dfrac{-1}{2015}\)
1)\(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|=\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
Cái này tự tính được nhé
2) \(\dfrac{6}{x+1}.\dfrac{x-1}{3}\in Z\Leftrightarrow\dfrac{6\left(x-1\right)}{3\left(x+1\right)}\in Z\)
\(\Rightarrow6x-6⋮3x+3\)
\(\Rightarrow6x+6-12⋮3x+3\)
\(\Rightarrow12⋮3x+3\)
Ok:>
Câu 1:
Thay \(a=\dfrac{1}{2015}\) vào biểu thức \(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|\) ta được:
\(\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\)
\(=\left|\dfrac{2014}{4058210}-\dfrac{2015}{4058210}\right|+\left|\dfrac{2016}{4062240}-\dfrac{2015}{4062240}\right|\)
\(=\left|\dfrac{2014-2015}{4058210}\right|+\left|\dfrac{2016-2015}{4062240}\right|\)
\(=\left|-\dfrac{1}{4058210}\right|+\left|\dfrac{1}{4062240}\right|\)
\(=\dfrac{1}{4058210}+\dfrac{1}{4062240}\)
\(=\dfrac{1008}{4090695680}+\dfrac{1007}{4090695680}\)
\(=\dfrac{1008+1007}{4090695680}\)
\(=\dfrac{2015}{4090695680}\)
\(=\dfrac{2015}{4090695680}\)
\(=\dfrac{1}{2030112}\)
Vậy giá trị của biểu thức P tại \(a=\dfrac{1}{2015}\) là \(\dfrac{1}{2030112}\)
\(B=\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+....+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{3}{2014}+1\right)+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)\)
\(B=\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+....+\dfrac{2017}{2014}+\dfrac{2017}{2015}+\dfrac{2017}{2016}\)
\(B=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
\(\dfrac{B}{A}=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}=2017\)
\(\dfrac{B}{A}=\dfrac{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\left(\dfrac{2015}{2}+\dfrac{2}{2}\right)+\left(\dfrac{2014}{3}+\dfrac{3}{3}\right)+...+\left(\dfrac{1}{2016}+\dfrac{2016}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Vậy \(\dfrac{B}{A}=2017\)
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 \(x=-2018\)
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!
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
Ta có:
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-2\left(1+\dfrac{1}{2}+...+\dfrac{1}{2014}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2015}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{2017}\right)\)
\(=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
Mà \(P=\dfrac{1}{1008}+\dfrac{1}{1009}+\dfrac{1}{1010}+...+\dfrac{1}{2015}\)
\(\Rightarrow S=P\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2016}=0^{2016}=0\)
Vậy \(\left(S-P\right)^{2016}=0\)
Ta có :
\(\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2014}=\left(1-\dfrac{1}{2015}\right)+\left(1-\dfrac{1}{2016}\right)+\left(1+\dfrac{2}{2014}\right)\) \(=\left(1+1+1\right)-\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)\)
\(=3-\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)\)
Dễ thấy : \(\left(\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{2}{2014}\right)>0\) vì \(\dfrac{1}{2015}>\dfrac{1}{2016}\)
Do đó \(\dfrac{2014}{2015}+\dfrac{2015}{2016}+\dfrac{2016}{2014}>3\)
~ Học tốt ~
@Lâm Gia Bảo lập luận sai --> đáp số đúng là sao?
\(\dfrac{2014}{2015}=1-\dfrac{2014}{2015}\)
\(\dfrac{2015}{2016}=1-\dfrac{1}{2016}\)
\(\dfrac{2016}{2014}=1+\dfrac{2}{2014}\)
công lại
\(VT=3+\left(\dfrac{1}{2014}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2014}-\dfrac{1}{2016}\right)\)
dễ dàng nhận ra
\(\left\{{}\begin{matrix}\dfrac{1}{2014}>\dfrac{1}{2015}\\\dfrac{1}{2014}>\dfrac{1}{2016}\end{matrix}\right.\) \(\Rightarrow VT>3\)
Thay a=\(\dfrac{1}{2015}\) vào P, ta có:
\(P=\left|\dfrac{1}{2015}-\dfrac{1}{2014}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\) (*)
Vì \(\left\{{}\begin{matrix}\dfrac{1}{2015}< \dfrac{1}{2014}\\\dfrac{1}{2015}>\dfrac{1}{2016}\end{matrix}\right.\) nên (*) \(\Rightarrow P=\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}=\dfrac{1}{2030112}\)
Vậy ...
\(P=\left|a-\dfrac{1}{2014}\right|+\left|a-\dfrac{1}{2016}\right|\\ =\left|\dfrac{1}{2014}-a\right|+\left|a-\dfrac{1}{2016}\right|\\ =\left|\dfrac{1}{2014}-\dfrac{1}{2015}\right|+\left|\dfrac{1}{2015}-\dfrac{1}{2016}\right|\\ =\dfrac{1}{2014}-\dfrac{1}{2015}+\dfrac{1}{2015}-\dfrac{1}{2016}\\ =\dfrac{1}{2014}-\dfrac{1}{2016}\\ =\dfrac{2}{2014\cdot2016}\\ =\dfrac{1}{2030112}\)