a) (x + 3)(x - 2) = 0
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}}\)
b) \(\left(2x-7\right)^2=25=5^2=\left(-5\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-7=5\\2x-7=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=6\\x=1\end{cases}}}\)
c) \(\left(1-3x\right)^3=-8=\left(-2\right)^3\)
\(\Rightarrow1-3x=-2\)
\(\Rightarrow3x=3\)
\(\Rightarrow x=1\)
(x+3)(x-2)=0
=>\(\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
(2x-7)2=25
(2x-7)2=52
=>2x-7=5
2x=5+7
2x=12
x=12:2
x=6
(1-3x)3=-8
(1-3x)3=(-2)3
=>1-3x=-2
3x=1-(-2)
3x=3
x=3:3
x=1
1) x = { -3; 2 }
2) ( 2x - 7 )2 = 25
( 2x - 7 )2 = 52
2x - 7 = 5
2x = 5 + 7
2x = 12
x = 12 : 2
x = 6
3) x = 1
a) \(\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0-3\\x=0+2\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}}\)
Vậy \(x\in\left\{-3;2\right\}\)
b) \(\left(2x-7\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-7\right)^2=5^2\\\left(2x-7\right)^2=\left(-5\right)^2\end{cases}\Rightarrow\orbr{\begin{cases}2x-7=5\\2x-7=-5\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x=1\end{cases}}}\)
Vậy \(x\in\left\{1;6\right\}\)
c) \(\left(1-3x\right)^3=-8\)
\(\left(1-3x\right)^3=\left(-2\right)^3\)
\(\Rightarrow1-3x=-2\)
\(3x=1-\left(-2\right)\)
\(3x=3\)
\(x=3\div3\)
\(x=1\)