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\(a,=\left[x^2\left(x^2-x-1\right)+x^3+x^2-3x-1\right]:\left(x^2-x-1\right)\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2x^2-2x-1\right]\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)\\ =\left[\left(x^2+x+2\right)\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)=x^2+x+2R1\)
a) \(\left(6x^2y-\dfrac{1}{2}xy+12y\right)\left(-\dfrac{1}{3}xy\right)=-2x^3y^2+\dfrac{1}{6}x^2y^2-4xy^2\)
b) \(\left(2x+3-y\right)\left(2x-y\right)=4x^2+6x-2xy-2xy-3y+y^2=4x^2+y^2+6x-3y-4xy\)
c) \(3\left(4x+1\right)\left(4x-1\right)-12\left(4x^2+1\right)=3\left(16x^2-1\right)-48x^2-12=48x^2-3-48x^2-12=-15\)
b. (2x + 3 - y)(2x - y)
= 4x2 - 2xy + 6x - 3y - 2xy + y2
= 4x2 - 4xy + 6x - 3y + y2
= \(\left[\left(2x\right)^2-4xy+y^2\right]\) + (6x - 3y)
= (2x - y)2 + 3(2x - y)
= (2x - y + 3)(2x - y)
Tham khảo
a)
-7x2(3x - 4y)
= -7x2.3x + 7x2ư.4y
= -21x2 + 28x2y
b)
(x - 3)(5x - 4)
= x.5x - x.4 - 3.5x + 3.4
= 5x2 - 4x - 15x + 12
= 5x2 - 19x + 12
c)
(2x - 1)2 = 4x2 - 4x + 1
d)
(x + 3)(x - 3) = x2 - 32 = x2 - 9
Lời giải:
a.
$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$
$=4x^2-12x+9+4x^2-4x-15$
$=24-8x$
b.
$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$
c.
$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$
$=6x^2-24x-6x^2-32x+10$
$=8x-10$
d.
$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$
$=x^2-9-x^2+10x-25=10x-34$
e.
$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$
$=-3x^2y+3xy^2=3xy(y-x)$
a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)
\(=4x^2-12x+9+2x-4x^2+15-6x\)
\(=-16x+24\)
b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)
\(=6x-9+5x+10\)
\(=11x+1\)
c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)
\(=6x^2-24x+30x-6x^2-10+2x\)
\(=8x-10\)