tìm x biet
a. (x-2)tất cả mũ 20 - (x-2) tất cả mũ 18 =0.x
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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
Rút gọn hả bạn ?
( 3x - 1 )2 - 9( x - 1 )( x + 1 )
= 9x2 - 6x + 1 - 9( x2 - 1 )
= 9x2 - 6x + 1 - 9x2 + 9
= 10 - 6x
( 2x + 3 )( 2x - 3 ) - ( 2x - 1 )2 - ( x - 1 )
= 4x2 - 9 - ( 4x2 - 4x + 1 ) - x + 1
= 4x2 - x - 8 - 4x2 + 4x - 1
= 3x - 9
2( x - 2y )( x + 2y ) + ( x - 2y )2 + ( x + 2y )2
= [ ( x + 2y ) + ( x - 2y ) ]2
= [ x + 2y + x - 2y ]2
= ( 2x )2 = 4x2
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) \(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)
a, \(\left(x+\frac{4}{3}y^2\right)^2\)
\(=x^2+\frac{8}{3}xy^2+\frac{16}{9}y^4\)
b, \(\left(2x-3y\right)^2\)
\(=4x^2-12xy+9y^2\)
c, \(\left(x^2+2x\right)\left(2x-x^2\right)\)
\(=\left(2x+x^2\right)\left(2x-x^2\right)\)
\(=4x^2-x^4\)
d, \(\left(x+\frac{1}{2}\right)^3\)
\(=x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
e, \(\left(2x-\frac{6}{5}y\right)^3\)
\(=8x^3-\frac{72}{5}x^2y+\frac{216}{25}xy^2-\frac{216}{125}y^3\)
(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