\(a,78-\left(x-3\right)=16\)
\(78-x+3=16\)
\(51-x=16\)
\(x=51-16=35\)
\(b,\left(2x+3\right):4=60\)
\(2x+3=60\cdot4=240\)
\(2x=240-3=237\)
\(x=\frac{237}{2}\)
\(c,\left(2x-8\right)=\left(2x-8\right)^7\)
\(\left(2x-8\right)-\left(2x-8\right)^7=0\)
Để mik nghĩ thêm tí r làm tiép
Để tui làm nốt ý C cho !!!
( 2x - 8 ) = ( 2x - 8 )7
=> ( 2x - 8 ) . 1 - ( 2x - 8 ) . ( 2x - 8 )6 = 0
=> ( 2x - 8 ) . [ 1 - ( 2x - 8 )6 ] = 0
=> 1 - ( 2x - 8 )6 = 0
=> ( 2x - 8 )6 = 1
=> ( 2x - 8 )6 = ( 16 )
\(\Rightarrow\orbr{\begin{cases}2x-8=1\\2x-8=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=9\\2x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{7}{2}\end{cases}}}\)
Vậy ........
c) \(\left(2x-8\right)=\left(2x-8\right)^7\)
\(\Leftrightarrow\left(2x-8\right)^7-\left(2x-8\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left[\left(2x-8\right)^6-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-8=0\\\left(2x-8\right)^6-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\\left(2x-8\right)^6=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x\in\left\{\frac{7}{2};\frac{9}{2}\right\}\end{cases}}\)
Vậy \(x\in\left\{4;\frac{7}{2};\frac{9}{2}\right\}\)
a,78−(x−3)=16
78−x+3=16
51−x=16
x=51−16=35
b,(2x+3):4=60
2x+3=60·4=240
2x=240−3=237
x=237/2
c,(2x−8)=(2x−8)7
(2x -8) - (2x -8^7)=0
