Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) A = (x - 3)(x² + 3x + 9) - (x³ + 3)
= x³ - 3³ - x³ - 3
= (x³ - x³) + (-27 - 3)
= -30
b) B = (2x + 1)(4x² - 2x + 1) - 8(x + 1/2)(x² - 1/2 x + 1/4)
= (2x)³ + 1³ - 8[x³ + (1/2)³]
= 8x³ + 1 - 8(x³ + 1/8)
= 8x³ + 1 - 8x³ - 1
= (8x³ - 8x³) + (1 - 1)
= 0
a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)
=3x-2-2x^2+2x-5x+5
=-2x^2+3
b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)
c: =x^3-3x^2+3x-1-x^3-1+9x^2-1
=6x^2+3x-3
\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)
\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)
\(=-2x^2+3\)
\(b,\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)\)
\(=\left(2x+1\right)\left[\left(2x+1\right)-2\left(3-x\right)\right]\)
\(=\left(2x+1\right)\left(2x+1-6+2x\right)\)
\(=\left(2x+1\right)\left(4x-5\right)\)
\(c,\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\)
\(=x^3-3x^2+3x-1-x^3-1-\left(3x-9x^2+1-3x\right)\)
\(=-3x^2+3x-2-3x+9x^2-1+3x\)
\(=6x^2+3x-3\)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
a) A = (x + 2)³ + (x - 2)³ - 2x(x² + 12)
= x³ + 6x² + 12x + 8 + x³ - 6x² + 12x - 8 - 2x² - 24x
= (x³ + x³) + (6x² - 6x² - 2x²) + (12x + 12x - 24x) + (8 - 8)
= 2x³ -2x²
b) B = (xy + 2)³ - 6(xy + 2)² + 12(xy + 2) - 8
= (xy + 2 - 2)³
= (xy)³
= x³y³
a: (1-2x)^3-(1+2x)^3
\(=1^3-3\cdot1^2\cdot2x+3\cdot1\cdot\left(2x\right)^2-8x^3-8x^3-12x^2-6x-1\)
\(=1-6x+12x^2-8x^3-8x^3-12x^2-6x-1\)
\(=-16x^3-12x\)
b: \(=x^3-6x^2+12x-8-x^3-x^2+8\)
\(=-7x^2+12x\)
c: \(=x^3+8-12x+6x^2-x^3+6x^2+12x\)
\(=12x^2+8\)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
a: Ta có: \(\left(x+1\right)^2+\left(x-1\right)^2-2\left(1+x\right)\left(1-x\right)\)
\(=\left(x+1\right)^2+2\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(x+1+x-1\right)^2\)
\(=4x^2\)
c: Ta có: \(3\left(x+2\right)^2-\left(3x+1\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=3x^2+12x+12-3x^2-16x-5+x^2+10x+25\)
\(=x^2+6x+32\)
a: =x^3+6x^2+12x+8-(x^3+3x^2+3x+1)
=x^3+6x^2+12x+8-x^3-3x^2-3x-1
=3x^2+9x+7
b: =x^3-9x^2+27x-27-x(x^2-6x+9)
=x^3-9x^2+27x-27-x^3+6x^2-9x
=-3x^2+18x-27
c: =x^3+3x^2+3x+1-x^3-x^2-2x^2-4x
=-x+1
\(a,\left(x+2\right)^3-\left(x+1\right)^3\\ =\left(x+2-x-1\right)\left(x^2+4x+4+x^2+3x+2+x^2+2x+1\right)\\ =3x^2+9x+7\\ b,\left(x-3\right)^3-x\left(x-3\right)^2\\ =x^3-6x^2+9x-27-x^3+6x^2-9x\\ =-27\)