tìm x biết \(\dfrac{1}{2016}\):2015x=-\(\dfrac{1}{2015}\)
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VM
Vũ Minh Tuấn
16 tháng 2 2020
Đúng(1)
Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 11 2018
\(\left(x^2+2015x\right)\left(\dfrac{1}{2016}+\dfrac{1}{1008}+\dfrac{1}{672}+1\right)=2022\)
\(\Leftrightarrow\left(x^2+2015x\right).\dfrac{2022}{2016}=2022\)
\(\Leftrightarrow x^2+2015x=2016\)
\(\Leftrightarrow x^2+2015x-2016=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2016\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2016\end{matrix}\right.\)
NG
23 tháng 1 2021
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
NH
5 tháng 3 2023
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
14 tháng 7 2017
Các câu dễ tự làm :v
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1981}=-4\) (sau khi đã sửa đề)
\(\Rightarrow\left(\dfrac{45-x}{1968}+1\right)+\left(\dfrac{40-x}{1973}+1\right)+\left(\dfrac{35-x}{1978}+1\right)+\left(\dfrac{30-x}{1981}+1\right)=0\)\(\Rightarrow\dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1981}=0\)
\(\Rightarrow\left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1981}\right)=0\)
\(\Rightarrow2013-x=0\Rightarrow x=2013\)
\(1+5+9+13+17+.....+x=5050\)
Số các số hạng là:
\(\dfrac{x-1}{4}+1=\dfrac{1}{4}x+\dfrac{3}{4}\)
Như vậy có :
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\) số hạng
Theo đề bài ta có:
\(\left(\dfrac{1}{4}x+\dfrac{3}{4}\right):2\left(x+1\right)=5050\)
\(\Rightarrow\left(\dfrac{1}{4}x+\dfrac{3}{4}\right)\left(x+1\right)=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+\dfrac{1}{4}x+\dfrac{3}{4}x+\dfrac{3}{4}=10100\)
\(\Rightarrow\dfrac{1}{4}x^2+x+\dfrac{3}{4}=10100\)
Kiệt sức.đến đây ko nghĩ nổi nx
MV
14 tháng 7 2017
a,
\(5^x+5^{x+2}=650\\ 5^x\left(1+5^2\right)=650\\ 5^x\cdot26=650\\ 5^x=25\\ 5^x=5^2\\ \Rightarrow x=2\)
Vậy \(x=2\)
b,
\(\left(x+2\right)^2=81\\ \Rightarrow\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
Vậy \(x=7\) hoặc \(x=-11\)
d,
\(\dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}=-4\\ \dfrac{45-x}{1968}+\dfrac{40-x}{1973}+\dfrac{35-x}{1978}+\dfrac{30-x}{1983}+4=0\\ \dfrac{45-x}{1968}+1+\dfrac{40-x}{1973}+1+\dfrac{35-x}{1978}+1+\dfrac{30-x}{1983}+1=0\\ \dfrac{2013-x}{1968}+\dfrac{2013-x}{1973}+\dfrac{2013-x}{1978}+\dfrac{2013-x}{1983}=0\\ \left(2013-x\right)\left(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\right)=0\)
Vì \(\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}+\dfrac{1}{1983}\ne0\) nên
\(2013-x=0\\ x=2013\)
Vậy \(x=2013\)
e,
\(\dfrac{1}{2016}:2015x=\dfrac{-1}{2015}\\ 2015x=\dfrac{-2015}{2016}\\ x=\dfrac{-1}{2016}\)
Vậy \(x=\dfrac{-1}{2016}\)
4 tháng 2 2021
a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)
\(=\dfrac{1}{2016}\)
b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)
\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)
\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)
hay x=10
Vậy: x=10
TH
3
MN
29 tháng 3 2020
P(x) = x2016 - 2015x2015 - 2015x2014 - ... - 2015x2 - 2015x
<=> P(x) = x2016 - 2016x2015 + x2015 - 2016x2014 + x2014 - ... - 2016x2 + x2 - 2016x + x
<=> P(2016) = 20162016 - 2016.20162015 + 20162015 - 2016.20162014 + 20162014 -...- 2016.20162 + 20162 - 2016.2016 + 2016
<=> P(2016)=20162016 - 20162016 + 20162015 - 20162015 + 20162014 - ... - 20163 + 20162 - 20162 + 2016
<=> P(2016) = 2016
Vậy P(2016) = 2016
29 tháng 3 2020
Ta có:
P(2016) = 20162016 - 2015 . 20162015 - 2015 . 20162014 -.....- 2015 . 20162 - 2015 . 2016 - 1
P(2016) = 20162016 - ( 2016 - 1 ) . 20162015 - ( 2016 -1 ) . 20162014 - ..... - ( 2016 - 1 ) . 20162 - ( 2016 - 1 ) . 2016 - 1
P(2016)= 20162016 - 20162016 + 20162015 - 20162015 + 20162014 - ..... - 20163 + 20162 - 20162 + 2016 - 1
P(2016) = 2016 - 1
P(2016) = 2015.
16 tháng 8 2021
Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)
Khi đó phương trình trở thành:
\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)
Tick plz
4 tháng 9 2022
=>|x-1|+|x-2|=2016
TH1: x<1
Pt sẽ là 1-x+2-x=2016
=>-2x+3=2016
=>-2x=2013
=>x=-2013/2(nhận)
TH2: 1<=x<2
Pt sẽ là x-1+2-x=2016
=>1=2016(loại)
TH3: x>=2
Pt sẽ là 2x-3=2016
=>2x=2019
=>x=2019/2(nhận)