a.\(\left(x+2\right)^2=81\)
b.\(5^x+5^{x+2}=650\)
c,\(\left(x-7\right)^{x+1}-\left(x-1\right)^{x+11}=0\)
d.\(\left(2x-1\right)^2-5=20\)
e.\(\left(x+2\right)^2=\frac{1}{2}-\frac{1}{3}\)
g.\(\left(x-1\right)^3=\left(x-1\right)\)
h.\(\left(x-1\right)^{x+2}=\left(x-1\right)^2\)
Các bạn mk vs nhé mk đg cần gấp
Thank you
a. +) x+2=9 +) x+2=-9
=> x=7 =>x=-11
a) (x + 2)2 = 81
=> (x + 2)2 = 92
=> \(\orbr{\begin{cases}x+2=-9\\x+2=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\x=7\end{cases}}\)
b) 5x + 5x + 2 = 650
=> 5x + 5x . 52 = 650
=> 5x + 5x . 25 = 650
=> 5x (25 + 1) = 650
=> 5x . 26 = 650
=> 5x = 650 : 26
=> 5x = 25
=> 5x = 52
=> x = 2
d) (2x - 1)2 - 5 = 20
=> (2x - 1)2 = 25
=> (2x - 1)2 = 52
=> \(\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\2x=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
g) (x - 1)3 = (x - 1)
=> (x - 1)3 - (x - 1) = 0
=> (x - 1) .[(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}x-1=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x-1=\pm1\end{cases}}}\)
Nếu x - 1 = 1
=> x = 2
Nếu x - 1 = -1
=> x = 0
Vậy \(x\in\left\{0;1;2\right\}\)