Bài 1:

|x-2|=4-x

ĐK: \(4-x\ge0\Leftrightarrow x\le4\)

Ta có: \(\orbr{\begin{cases}x-2=4-x\\x-2=x-4\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\0=2\left(loại\right)\end{cases}\Rightarrow}}x=3\left(tm\right)\)

Vậy x = 3 

Bài 2:

a, sao có z

b, Vì \(\hept{\begin{cases}\left|2017-x\right|\ge0\\\left|y-x+2018\right|\ge0\end{cases}\Rightarrow\left|2017-x\right|+\left|y-x+2018\right|\ge0}\)

Mà |2017-x|+|y-x+2018|=0

\(\Rightarrow\hept{\begin{cases}\left|2017-x\right|=0\\\left|y-x+2018\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\y-2017+2018=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2017\\y=1\end{cases}}}\)

Vậy x=2017,y=1

c, giống b

Bài 2 cũng có z bạn ạ Làm luôn hộ mình câu b

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

cac ban oi ai xong truoc mk k cho nhe

a)\(x^{2016}=x^{2017}\)

 \(\Leftrightarrow x^{2017}-x^{2016}=0\)

\(\Leftrightarrow x^{2016}.\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^{2016}=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vay ...

b) \(2y.\left(x+1\right)-x-7=0\)

\(\Leftrightarrow2y.\left(x+1\right)-\left(x+1\right)=6\)

\(\Leftrightarrow\left(x+1\right).\left(2y+1\right)=6\)

Đến chỗ này bạn tự tìm các cặp x,y nha

( 3x - 2⁴ ) . 7⁵ = 2.7⁶ .1/2017⁰

( 3x - 2⁴ ) . 7⁵ =235298

( 3x - 2⁴ ) = 235298 : 7⁵ 

( 3x - 2⁴ ) =14

3x = 14 + 2⁴ 

3x = 30

x = 0

dung 100%

\(\left(3x-2^4\right).7^5=2.7^6.\frac{1}{2017^0}\)

\(\Leftrightarrow\left(3x-16\right).7^5=2.7^6.1\)

\(\Leftrightarrow3x-16=\frac{2.7^6}{7^5}\)

\(\Leftrightarrow3x-16=2.7\)

\(\Leftrightarrow3x-16=14\)

\(\Leftrightarrow3x=30\)

\(\Leftrightarrow x=10\)