-x+2016-x+2016=0

-2x-4032=0

-2x=4032

x=-2016

câu b tương tự nhé Vương Tuấn Khải

a,x=2016

b,x=-2015

Các bn ghi rõ cách làm ra đi

|(x+2014)(x-2015)| + |(x-2015)(x+2016)| = 0

<=> |x²+2014x-2015x-2014.2015| + |x²-2015x+2016x-2015.2016| = 0

<=> |x²-x-2014.2015| + |x²+x-2015.2016| = 0

\(\Leftrightarrow\left\{\begin{matrix}x^2-x-2014.2015=0\\x^2+x-2015.2016=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\left(x-1\right)=2014.2015\\x\left(x+1\right)=2015.2016\end{matrix}\right.\)<=> x=2015

Vậy ...

\(\left(x-2015\right)\left(x+2016\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2015=0\\x+2016=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2015\\x=-2016\end{matrix}\right.\)

vậy ...

\(\left(x-2015\right)\left(x+2016\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2015=0\\x+2016=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2015\\x=-2016\end{matrix}\right.\)

Vậy \(x\in\left\{2015;-2016\right\}\)

 Mình nhầm \(C^1_{2016}a_{2015}\)thành  \(C^1_{2016}a^{2015}\)

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3