Thực hiện phép tính sau
(2x-3)^2-x(4x-13)-12x^3÷6x^2
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a, Biến đổi vế trái :
\(VT=x\left(x+1\right)\left(x+2\right)=\left(x^2+x\right)\left(x+2\right)=x^3+3x^2+2x\) 2x
b,\(\left(3x-2\right)\left(4x-5\right)-\left(2x-1\right)\left(6x+2\right)=0\)
\(\Leftrightarrow12x^2-15x-8x+10-\left(12x^2+4x-6x-2\right)=0\)
\(\Leftrightarrow12x^2-23x+10-12x^2+2x+2=0\)
\(\Leftrightarrow12-21x=0\)
\(\Leftrightarrow-21x=-12\)
\(\Leftrightarrow21x=12\)
\(\Leftrightarrow x=\frac{4}{7}\)
c,
a: =x^4-3x^5+4x^8
b: =2x^3+2x^2+4x
c: =4x^2+8x-5
d: =2x+3x^2+7x^4
a) Ta có P = ( 4 x 2 − 1 ) ( 2 x + 1 ) − ( 2 x − 1 ) − ( 4 x 2 − 1 ) ( 2 x + 1 ) ( 2 x − 1 ) = 3 − 4 x 2
b) Ta có Q = 3 x ( x + 3 ) . ( x + 3 ) ( x − 3 ) − x = 9 − 3 x x + 3
a) 3x2 .(2x2 - 3yz + x3)= 6x4 - 6x2yz +3x5
b)(24x5 - 12x4 + 6x2 ).6x2 = 144x7 - 72x6 +36x4
a) 3x2 . (2x2 - 3yz + x3)
= 3x2 . 2x2 + 3x2 . (- 3yz) + 3x2 . x3
= 6x4 + (-9x2yz) + 3x5
= 6x4 - 9x2yz + 3x5
a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)
=-4/3x^2+8/3-10/3
=-4/3x^2-2/3
d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)
\(=3x^2+9x+22+\dfrac{68}{x-3}\)
a) ( 3x - 2 )( 4x + 5 ) - 6x( 2x - 1 )
= 12x2 + 7x - 10 - 12x2 + 6x
= 13x - 10
b) ( 2x - 5 )2 - 4( x + 3 )( x - 3 )
= 4x2 - 20x + 25 - 4( x2 - 9 )
= 4x2 - 20x + 25 - 4x2 + 36
= 61 - 20x
c) 2x3 - 5x2 + 7x - 6
= 2x3 - 3x2 - 2x2 + 3x + 4x - 6
= x2( 2x - 3 ) - x( 2x - 3 ) + 2( 2x - 3 )
= ( 2x - 3 )( x2 - x + 2 )
=> ( 2x3 - 5x2 + 7x - 6 ) : ( 2x - 3 ) = x2 - x + 2
a, (3x - 2 ) (4x + 5) - 6x (2x -1) = ( 7x + 15x -8x - 10 ) - ( 12x2 -6x ) = 7x2 + 15x - 8x -10 -12x2 + 6x = -5x2 + x - 10
\(=4x^2-12x+9-4x^2+13x-2x=-x+9\)
\(= 4x^2 - 12x + 9 - 4x^2 + 13x - 2x = -x + 9\)