K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

HQ
Hà Quang Minh
Giáo viên
20 tháng 9 2023

\(\begin{array}{l}(4x - 3)(2{x^2} + 5x - 6)\\ = 4x.2{x^2} + 4x.5x - 6.4x - 3.2{x^2} - 3.5x + 18\\ = 8{x^3} + 20{x^2} - 6{x^2} - 24x - 15x + 18\\ = 8{x^3} + 14{x^2} - 39x + 18\end{array}\) 

2 tháng 5 2023

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

2 tháng 5 2023

a, \(4x^2\left(x^2-5x+2\right)\\ =4x^4-20x^3+8x^2\)

b, \(\left(2x^2-5x+3\right):\left(2x-3\right)\\ =x-1\)

a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)

b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)

c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)

d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

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: \(\dfrac{16x^3y^4}{4x^2y^2}=4xy^2\)

27 tháng 10 2021

a: \(=15x^4-12x^3+9x^2\)

c: \(=5x^3-15x^2-4x^2+12x\)

\(=5x^3-19x^2+12x\)

27 tháng 10 2021

còn câu b và d sao ak

 

 

Bài 3:

3: \(6x\left(x-y\right)-9y^2+9xy\)

\(=6x\left(x-y\right)+9xy-9y^2\)

\(=6x\left(x-y\right)+9y\left(x-y\right)\)

\(=\left(x-y\right)\left(6x+9y\right)\)

\(=3\left(2x+3y\right)\left(x-y\right)\)

Bài 4:

loading...

loading...

loading...

13 tháng 12 2023

ĐKXĐ: \(x\notin\left\{2;-3;-4\right\}\)

\(\dfrac{x^2-5x+6}{x^2+7x+12}\cdot\dfrac{x^2+3x}{x^2-4x+4}\)

\(=\dfrac{x^2-2x-3x+6}{x^2+3x+4x+12}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)

\(=\dfrac{x\left(x-2\right)-3\left(x-2\right)}{x\left(x+3\right)+4\left(x+3\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)

\(=\dfrac{\left(x-3\right)\left(x-2\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)

\(=\dfrac{x\left(x-3\right)}{\left(x-2\right)\left(x+4\right)}\)

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: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)