Thực hiện phép tính:
a) (2x^3-4x^2+5x):(-2x)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a.
$5x-[2x+1-(2x-3)-(4x+1)]=5x-(2x+1-2x+3-4x-1)$
$=5x-(-4x+3)=5x+4x-3=9x-3$
b.
$(-3x^2+2x-1)+(4x^2-2x+3)$
$=-3x^2+2x-1+4x^2-2x+3=x^2+2$
a) 4x²(x² - 5x + 2)
= 4x².x² - 4x².5x + 4x².2
= 4x⁴ - 20x³ + 8x²
b) (2x² - 5x + 3) : (2x - 3)
= (2x² - 3x - 2x + 3) : (2x - 3)
= [(2x² - 3x) - (2x - 3)] : (2x - 3)
= [x(2x - 3) - (2x - 3)] : (2x - 3)
= (2x - 3)(x - 1) : (2x - 3)
= x - 1
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: \(\left(5x-3\right)\left(5x+3\right)=25x^2-9\)
b: \(=2x^2-12x^3+15x^2-18x=-12x^3+17x^2-18x\)
c: \(\dfrac{16x^3y^4}{4x^2y^2}=4xy^2\)
a: \(\dfrac{4-x^2}{x-3}+\dfrac{2x-2x^2}{3-x}+\dfrac{5-4x}{x-3}\)
\(=\dfrac{4-x^2-2x+2x^2+5-4x}{x-3}=\dfrac{x^2-6x+9}{x-3}\)
=(x-3)^2/(x-3)
=x-3
b: \(\dfrac{2}{x+2}+\dfrac{-4}{2-x}+\dfrac{5x+2}{4-x^2}\)
\(=\dfrac{2}{x+2}-\dfrac{4}{x-2}-\dfrac{5x+2}{x^2-4}\)
\(=\dfrac{2x-4-4x-8-5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-7x-14}{\left(x-2\right)\left(x+2\right)}\)
=-7(x+2)/(x-2)(x+2)
=-7/(x-2)
a) –4x(5x^2 – 2xy + y^2)
= -20x^3 + 8x^2y - 4xy^2
b) (4x – 1)(2x^2 – x – 1)
= 8x^3 - 4x^2 - 4x - 2x^2 + x + 1
= 8x^3 - 6x^2 - 3x + 1