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a.(3^2+4^2).x=10^2
(9+16).x =100
25.x =100
x =100:25
x =4
b.(x-5)^2 =81
x-5 =9
x =9+5
x =14
c.(2x+1)^3 = 343
2x+1 = 7
2x =7-1
2x =6
x =6:2
x = 3
1) \(\Rightarrow x^2\left(x^{2004}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{2004}=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
2) \(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
\(7\left(5-x\right)+5\left(x-2\right)=15\)
\(3x-7x+5x-10=15\)
\(x-10=15\)
\(x=25\)
1, x-(-19)+(-2)5=14-(-2)4
x+19+(-32)=14-16
x+19+(-32)=-2
x+19=-2-(-32)
x+19=30
x=30-19
x=11
Vậy x=11
2)7.(5-x)+5(x-2)=15
35-7x+5x-10=15
-7x+5x=15-35+10
-2x=-10
2x=10
x=10:2
x=5
Vậy x=5
Ta có: \(\left(x-2\right)^2-25=0\)
\(\Leftrightarrow\left(x-2\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{7;-3\right\}\)
(x - 2)20 - (x - 2)18 = 0
=> (x - 2)18 [ (x - 2)2 - 1 ] = 0
=> (x - 2)18 = 0 => x - 2 = 0 => x = 2
hoặc (x - 2)2 - 1 = 0 => (x - 2)2 = 1 => x - 2 = 1 => x = 3
Vậy x = 2 ; x = 3