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\(3x\left(x^2+2y\right)^2-12xy\left(x^2+y\right)\)
\(=3x\left(x^4+4x^2y+4y^2\right)-12xy\left(x^2+y\right)\)
\(=3x^5+12x^3y+12xy^2-12xy\left(x^2-y\right)\)
\(=3x^5+\left(12x^3y\right)^2-\left(12xy^2\right)^2\)
\(=3x^5\)
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> 18x=36
=> x=2
Vậy x=2
Team 2k5 đúng ko, k mk nha !!
7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)
\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)
\(\Leftrightarrow11x=4\)
hay \(x=\dfrac{4}{11}\)
8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)
\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)
\(\Leftrightarrow x^2=9\)
hay \(x\in\left\{3;-3\right\}\)
Bài 1.
a) -2x( -3x + 2 ) - ( x + 2 )2
= 6x2 - 4x - ( x2 + 4x + 4 )
= 6x2 - 4x - x2 - 4x - 4
= 5x2 - 8x - 4
b) ( x + 2 )( x2 - 2x + 4 ) - 2( x + 1 )( 1 - x )
= x3 + 8 + 2( x + 1 )( x - 1 )
= x3 + 8 + 2( x2 - 1 )
= x3 + 8 + 2x2 - 2
= x3 + 2x2 + 6
c) ( 2x - 1 )2 - 2( 4x2 - 1 ) + ( 2x + 1 )2
= 4x2 - 4x + 1 - 8x2 + 2 + 4x2 + 4x + 1
= 4
d) x2 - 3x + xy - 3y
= x( x - 3 ) + y( x - 3 )
= ( x - 3 )( x + y )
Bài 2.
a) 4x2 - 4xy + y2 = ( 2x - y )2
b) 9x3 - 9x2y - 4x + 4y
= 9x2( x - y ) - 4( x - y )
= ( x - y )( 9x2 - 4 )
= ( x - y )( 3x - 2 )( 3x + 2 )
c) x3 + 2 + 3( x3 - 2 )
= x3 + 2 + 3x3 - 6
= 4x3 - 4
= 4( x3 - 1 )
= 4( x - 1 )( x2 + x + 1 )
Bài 3.
2( x - 2 ) = x2 - 4x + 4
⇔ ( x - 2 )2 - 2( x - 2 ) = 0
⇔ ( x - 2 )( x - 2 - 2 ) = 0
⇔ ( x - 2 )( x - 4 ) = 0
⇔ x = 2 hoặc x = 4
1: Khi x=2 thì \(A=\dfrac{4\cdot2+1}{2-1}=9\)
2: \(=\dfrac{3x+1-2x^2-2x+3x^2-3x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
