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,\(x\in\left\{-2010;-2022\right\}\)
Theo đề bài ta có :
(x+2016) . (x+2016) = -3 . (-12)
(x+2016)2 = 36
Ta có (x+2016)2= 62
Trường hợp 1:
x+2016=6
x = 6-2016
x = -2010
Trường hợp 2:
x+2016= -6
x = -6-2016
x = -2022
Vậy x = -2010; x= -2022
k mk nhé
suy ra (x+2016).(x+2016)=(-3).(-12)
(x+2016)^2 =36
x+2016=6 suy ra x=-2010
hoặc x+2016=-6 suy ra x=-2022
vậy x=-2010 hoặc x=-2022
nhân tích chéo ta được
(x+2006).(x+2006)=(-3).(-12)
(x+2006)^2=36
hay(x+2006)^2=6^2
x+2006=6
x=6-2006
x=-2000
