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4x(x2-2x+3)-3x(x+1)(x-2)+(2x-5)(x-7)-4x(x2-x+1)
=4x3-8x2+12x-3x(x2-2x+x-2)+2x2-14x-5x+35-4x3+4x2-4x
=4x3-8x2+12x-3x3+6x2-3x2+6x+2x2-14x-5x+35-4x3+4x2-4x
=-3x3+x2-5x+35
#H
1.
a. x3 - 4x2 - xy2 + 4x
= x ( x2 - 4x + 4 - y2 )
= x [ ( x - 2 )2 - y2 ]
= x ( x - y - 2 ) ( x + y - 2 )
b. x2 - x - 2 = x2 + x - 2x - 2 = x ( x + 1 ) - 2 ( x + 1 ) = ( x - 2 ) ( x + 1 )
c. x4 + 4
= ( x4 + 2x3 + 2x2 ) - ( 2x3 + 4x2 + 4x ) + ( 2x2 + 4x + 4 )
= x2 ( x2 + 2x + 2 ) - 2x ( x2 + 2x + 2 ) + 2 ( x2 + 2x + 2 )
= ( x2 + 2x + 2 ) ( x2 - 2x + 2 )
Cái áo rank bàng mùa 11 phải ko xai đừng hcuiwr
a) \(8x^3-y^3-6xy\left(2x-y\right)=\left(2x-y\right)\left(4x^2+2xy+y^2\right)-6xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2-6xy\right)=\left(2x-y\right)\left(4x^2-4xy+y^2\right)\)
\(=\left(2x-y\right)\left(2x-y\right)^2=\left(2x-y\right)^3\)
b) \(\left(3x+2\right)^2-2\left(x-1\right)\left(3x+2\right)+\left(x-1\right)^2\)
\(=\left[\left(3x+2\right)-\left(x-1\right)\right]^2=\left(3x+2-x+1\right)^2=\left(2x+3\right)^2\)
a) 8x3 - y3 - 6xy(2x - y)
= (2x)3 - y3 - 3.2x.y.(2x - y)
= (2x - y)3
b) (3x + 2)2 - 2(x - 1)(3x + 2) + (x - 1)2
= (3x + 2 - x + 1)2
= (2x + 3)2
(3x - 7)(x + 3) + 5x(x2 - 2x - 4) - (4x - 5)(x-4)-(x + 2)(x + 1)
=3x2+9x-7x-21+5x3-10x2-20x-4x2+16x+5x-20-x2-x-2x-2
=5x3-13x2-43
-7x(2x2- 4x - 5) - (x - 5)(-2x + 3) + (3x - 2)(x+4) - 4x2 (x - 3)
=-14x3+28x2+35x+2x2-3x-10x+15+3x2+12x-2x-8-4x3+12x2
=-18x3+45x2+32x+7
Olm chào em, đề yêu cầu gì vậy em nhỉ?
\(\left(4-3x\right)^2-3\left(x+3\right)^2+\left(2x-1\right)\left(2x+1\right)\)
\(=9x^2-24x+16-3\left(x^2+6x+9\right)+4x^2-1\)
\(=13x^2-24x+15-3x^2-18x-27=10x^2-42x-12\)
Yêu cầu: Tính
\(\left(4-3x\right)^2-3\left(x+3\right)^2+\left(2x+1\right)\left(2x-1\right)\)
\(=16-24x+9x^2-3\left(x^2+6x+9\right)+\left(4x^2-1\right)\)
\(=16-24x+9x^2-3x^2-18x-27+4x^2-1\)
\(=\left(9x^2-3x^2+4x^2\right)+\left(-18x-24x\right)+\left(16-27-1\right)\)
\(=10x^2-42x-12\)
Yêu cầu: Giải phương trình
\(\left(4-3x\right)^2-3\left(x+3\right)^2+\left(2x-1\right)\left(2x+1\right)=0\)
\(10x^2-42x-12=0\)
\(5x^2-21x-6=0\)
\(5x^2-21x=6\)
\(x^2-\frac{21}{5}x=\frac65\)
\(x^2-2.x.\frac{21}{10}+\frac{441}{100}=\frac65+\frac{441}{100}\)
\(\left(x-\frac{21}{10}\right)^2=\frac{561}{100}\)
\(\left[\begin{matrix}x-\frac{21}{10}=\frac{\sqrt{561}}{10}\\ x-\frac{21}{10}=\frac{-\sqrt{561}}{10}\end{matrix}\right.\)
\(\left[\begin{matrix}x=\frac{21+\sqrt{561}}{10}\\ x=\frac{21-\sqrt{561}}{10}\end{matrix}\right.\)
Vậy \(S=\left\lbrace\frac{21\pm\sqrt{561}}{10}\right\rbrace.\)
(4-3x)^2-3(x+3)^2+(2x-1)(2x+1)
=16 - 24x + 9x^2 - 3(x^2 + 6x + 9) + 4x^2 - 1
=16 - 24x + 9x^2 - (3x^2 + 18x + 27) + 4x^2 - 1
=16 - 24x + 9x^2 - 3x^2 - 18x - 27 + 4x^2 - 1
=(9x^2 - 3x^2 + 4x^2) - (24x + 18x) + (16 - 27 - 1)
=10x^2 - 42x - 12