(x+2016)^2=-12.(-3)

(x+2016)^2=36

x+2016=6 hoặc x+2016=-6

x=-2010 ,x=-2022

trả lời

nhân chéo lên là làm đc nha bn

hok tốt

\(\frac{x+2016}{-3}=-\frac{12}{x+2016}\)

\(\Rightarrow\left(x+2016\right)^2=-3.\left(-12\right)\)

\(\Rightarrow\left(x+2016\right)^2=36\)

\(\Rightarrow\left(x+2016\right)^2=6^2\)

\(\Rightarrow x+2016=6\)

\(\Rightarrow x=6-2016\)

\(\Rightarrow x=-2010\)

\(\frac{x+2016}{-3}=\frac{-12}{x+2016}\)

\(\Leftrightarrow(x+2016)\cdot(x+2016)=(-12)\cdot(-3)\)

\(\Leftrightarrow(x+2016)^2=36\)

\(\Leftrightarrow(x+2016)^2=6^2\)

\(\Leftrightarrow x+2016=\pm6\)

\(\Leftrightarrow\hept{\begin{cases}x+2016=6\\x+2016=-6\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2010\\x=-2022\end{cases}}\)

Do đó,\(x\in\left\{-2010;-2022\right\}\)

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

\(\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1+\frac{x-3}{2014}-1+...+\frac{x-2016}{1}-1=0\)

\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}+...+\frac{x-2017}{1}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}+...+1\right)=0\)

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

Đáp án là : 

Tìm x : 

x = -2019 

x= -2019

ta có : 

\(\left(\frac{x}{3}-672\right)+\frac{\left(2016-x\right)}{13}+\frac{\left(x-2016\right)}{17}=0\)

hay \(\left(x-2016\right)\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{17}\right)=0\Leftrightarrow x=2016\)

\(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{3x+12}{2015}+3\)

\(\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{3 \left(x+2019\right)}{2015}\)

\(\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\right)=0\)

mà \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\ne0\Rightarrow x+2019=0\Leftrightarrow x=-2019\)

Ai biết làm giải giùm mình với

Bạn có chắc chắn là đúng không

trừ mỗi vế cho 2 rồi tách -2 thành -1và -1

X=1 nhé

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

\(\Leftrightarrow\)\(\frac{x+2014}{2015}-1+\frac{x+2015}{2016}-1=\frac{x+2016}{2017}-1+\frac{x+2017}{2018}-1\)

\(\Leftrightarrow\)\(\frac{x-1}{2015}+\frac{x-1}{2016}=\frac{x-1}{2017}+\frac{x-1}{2018}\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)   ( do  1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )

\(\Leftrightarrow\) \(x=1\)

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

\(2016x+x\frac{1}{2016}-2016=\frac{1}{2016}\)

\(\Rightarrow2016x-2016+x.\frac{1}{2016}-\frac{1}{2016}=0\)

\(\Rightarrow2016.\left(x-1\right)+\frac{1}{2016}.\left(x-1\right)=0\)

\(\Rightarrow\left(2016+\frac{1}{2016}\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2016+\frac{1}{1016}=0\text{ (loại vì }2016+\frac{1}{2016}>0\text{)}\text{ }\\x-1=0\end{cases}}\)

\(\Rightarrow x=1\)

\(2016x+x\frac{1}{2016}-2016=\frac{1}{2016}\)

\(\Leftrightarrow x\left(2016+\frac{1}{2016}\right)=\frac{1}{2016}+2016\)

\(\Leftrightarrow x=\left(2016+\frac{1}{2016}\right):\left(2016+\frac{1}{2016}\right)\)

\(\Leftrightarrow x=1\)

\(2016x+x\frac{1}{2016}-2016=\frac{1}{2016}\)

=> \(2016x+\frac{2016x+1}{2016}-2016=\frac{1}{2016}\)

=> \(2016x+x+\frac{1}{2016}-2016=\frac{1}{2016}\)

=> \(2017x-2016=\frac{1}{2016}-\frac{1}{2016}\)

=> \(2017x-2016=0\)

=> \(2017x=2016\)

=> \(x=2016:2017\)

=> \(x=\frac{2016}{2017}\)