Tìm x biết
\(\frac{x+2016}{-3}=\frac{-12}{x+2016}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{x+2016}{-3}=-\frac{12}{x+2016}\)
\(\Rightarrow\left(x+2016\right)^2=-3.\left(-12\right)\)
\(\Rightarrow\left(x+2016\right)^2=36\)
\(\Rightarrow\left(x+2016\right)^2=6^2\)
\(\Rightarrow x+2016=6\)
\(\Rightarrow x=6-2016\)
\(\Rightarrow x=-2010\)
\(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+...+\frac{x-2016}{1}=2016\)
\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+...+1\ne0\)
\(\Rightarrow x-2017=0\)
\(\Rightarrow x=2017\)
<=> \(\frac{x-1}{2016}+\frac{x-2}{2015}+\frac{x-3}{2014}+....+\frac{x-2016}{1}-2016=0\)\(=0\)
<=> \(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)+...+\left(\frac{x-2016}{1}-1\right)=0\)
<=> \(\frac{x-2017}{2016}+\frac{x-2017}{2015}+...+\frac{x-2017}{1}=0\)
<=> \(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}\right)=0\)
<=> \(x-2017=0\)\(\left(do\frac{1}{2016}+\frac{1}{2015}+...+\frac{1}{1}>0\right)\)
<=> \(x=2017\)
Vậy x = 2017
đúng thì
ta có :
\(\left(\frac{x}{3}-672\right)+\frac{\left(2016-x\right)}{13}+\frac{\left(x-2016\right)}{17}=0\)
hay \(\left(x-2016\right)\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{17}\right)=0\Leftrightarrow x=2016\)
Tìm \(x\varepsilonℝ\) biết \(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{3x+12}{2015}\)
\(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{3x+12}{2015}+3\)
\(\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{3 \left(x+2019\right)}{2015}\)
\(\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\right)=0\)
mà \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\ne0\Rightarrow x+2019=0\Leftrightarrow x=-2019\)
Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)
\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)
\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))
\(\Leftrightarrow x=1\)
Vạy x=1
\(2016x+x\frac{1}{2016}-2016=\frac{1}{2016}\)
\(\Rightarrow2016x-2016+x.\frac{1}{2016}-\frac{1}{2016}=0\)
\(\Rightarrow2016.\left(x-1\right)+\frac{1}{2016}.\left(x-1\right)=0\)
\(\Rightarrow\left(2016+\frac{1}{2016}\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2016+\frac{1}{1016}=0\text{ (loại vì }2016+\frac{1}{2016}>0\text{)}\text{ }\\x-1=0\end{cases}}\)
\(\Rightarrow x=1\)
\(2016x+x\frac{1}{2016}-2016=\frac{1}{2016}\)
\(\Leftrightarrow x\left(2016+\frac{1}{2016}\right)=\frac{1}{2016}+2016\)
\(\Leftrightarrow x=\left(2016+\frac{1}{2016}\right):\left(2016+\frac{1}{2016}\right)\)
\(\Leftrightarrow x=1\)
(x+2016)^2=-12.(-3)
(x+2016)^2=36
x+2016=6 hoặc x+2016=-6
x=-2010 ,x=-2022