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( 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
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
\(\left(2x^{2n}+3x^{2n-1}\right)\left(x^{1-2n}-3x^{2-2n}\right)=2x^{2n}\times x^{1-2n}-2x^{2n}\times3x^{2-2n}+3x^{2n-1}\times x^{1-2n}-3x^{2n-1}\times3x^{2-2n}\)
\(=2x-6x^2+3-3x=3-x-6x^2\)
Đặt x2n = a thì ta có
(\(2a+\frac{3a}{x}\))(\(\frac{x}{a}-\frac{3x^2}{a}\))
= (2x + 3)(1 - 3x) = 2x - 6x2 + 3 - 9x = - 6x2 - 7x + 3
\(\left(3x+2\right)^2+2x\left(1-2x\right)+4x\left(x-\dfrac{1}{2}\right)=9\)
\(\Leftrightarrow9x^2+12x+4+2x-4x^2+4x^2-2x-9=0\)
\(\Leftrightarrow9x^2+12x+4-9=0\)
\(\Leftrightarrow\left(3x+2\right)^2-3^2=0\)
\(\Leftrightarrow\left(3x+2-3\right)\left(3x+2+3\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{3}\) hoặc \(x=-\dfrac{5}{3}\)