-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

2016x(x - 2017/2016)=0

2016x^2-2017x=0

x(2016x-2017)=0

suy ra : x=0 hoặc 2016x - 2017 = 0

           x = 0 hoặc 2016x          =2017

           x = 0 hoặc x                 = 2017/2016

2) 5(x-2)+3x(2-x)=0

5x(x-2)-3x(2-x)=0

(x-2)(5--3x)=0

suy ra : x-2=0 hoặc 5-3x=0

           x=2  hoặc 3x   =5

          x=2 hoặc x =5/3

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

a, (x-2016)(2x-1)=0

<=>x=2016 hoặc x=-1/2

b, (x+2)(x+2-x+2)=0

<=>4(x+2)=0

<=>x+2=0

<=>x=-2

a) 2x(x-2016)-x+2016=0

=>2x(x-2016)-(x-2016)=0

=>(x-2016)(2x-1)=0

=>\(\left\{{}\begin{matrix}x-2016=0\\2x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2016\\x=\dfrac{1}{2}\end{matrix}\right.\)

vậy x=2016 hoặc x=\(\dfrac{1}{2}\)

b) (x+2)²-(x-2)(x+2)=0

=>(x+2)[(x+2)-(x-2)]=0

=>(x+2)(x+2-x+2)=0

=>(x+2)4=0

=>x+2=0

=>x=-2

vậy x=-2

Tìm x, biết :

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-2016=0\\x+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2016\\x=-1\end{matrix}\right.\)

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

\(\Leftrightarrow\left(x+1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2016=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2016\end{matrix}\right.\)

Vậy x=-1 hoặc x=2016.

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

\(\Leftrightarrow\left(x+1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2016=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2016\end{matrix}\right.\)

Vậy, ...

a... . x = 2016

b.....x=0

c.... x = -2

a.  x=2016

b.   x=2017

c.    x= -2

Các bạn giải rõ ràng hộ mình nha

6x.(x+2016) - (x+2016) = 0

=> (6x -1) (x +2016) =0

=> 2 trường hợp 

trường hợp 1 : 6x-1 = 0

=> x = 1/6

trường hợp 2 : x +2016 =0

=> x = - 2016

=> x € {1/6;-2016}

     \(6x\left(x+2016\right)-x-2016=0\)

\(6x\left(x+2016\right)-\left(x+2016\right)=0\)

                \(\left(x+2016\right)\left(6x-1\right)=0\)

=> \(x+2016=0\)hoặc \(6x-1=0\)

Với \(x+2016=0\)=>\(x=-2016\)

Với \(6x-1=0\)=>\(6x=1\)=>\(x=\frac{1}{6}\)

Vậy \(x=-2016;x=\frac{1}{4}\)

Ta có \(6x.\left(x+2016\right)-x-2016=0\)

\(\Rightarrow6x.\left(x+2016\right)-\left(x+2016\right)=0\)

\(\Rightarrow\left(x+2016\right).\left(1+6x\right)=0\)

\(\Rightarrow\hept{\begin{cases}2016+x=0\\1+6x=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-2016\\6x=-1\Rightarrow x=-\frac{1}{6}\end{cases}}\)

Vậy \(x\in\left\{-2016;-\frac{1}{6}\right\}\)