Những câu hỏi liên quan
Tuyết Ly
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Nguyễn Lê Phước Thịnh
14 tháng 5 2022 lúc 23:14

a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)

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

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

\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)

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

=1/3

d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)

\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)

Nguyen Anh Duc
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Nguyễn Lê Phước Thịnh
8 tháng 3 2023 lúc 23:05

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

Tuyết Ly
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Nguyễn Lê Phước Thịnh
14 tháng 5 2022 lúc 23:36

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)

Vũ Thảo Anh
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Nguyenngocdiem
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Nguyễn Lê Phước Thịnh
22 tháng 6 2023 lúc 16:22

a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y

=5x^3-7x^2y+2xy^2+5x-2y

b: =(x^2-1)(x+2)

=x^3+2x^2-x-2

c: =1/2x^2y^2(4x^2-y^2)

=2x^4y^2-1/2x^2y^4

d: =(x^2-1/4)(4x-1)

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

e: =x^2-2x-35+(2x+1)(x-3)

=x^2-2x-35+2x^2-6x+x-3

=3x^2-7x-38

Hân Hân
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Kirito-Kun
12 tháng 9 2021 lúc 19:28

2.

a. 3x(12x - 4) - 9x(4x - 3) = 30

<=> 36x2 - 12x - 36x2 + 27x = 30

<=> 36x2 - 36x2 - 12x + 27x = 30

<=> 15x = 30

<=> x = 2

b. x(5 - 2x) + 2x(x - 1) = 15

<=> 5x - 2x2 + 2x2 - 2x = 15

<=> -2x2 + 2x2 + 5x - 2x = 15

<=> 3x = 15

<=> x = 5

Phạm Trần Hoàng Anh
12 tháng 9 2021 lúc 19:30

a) x2 ( 5x3 - x - 2323x2y=  6969x3y2- 2323x4y+ 2323x2y2

c) x2 ( 4x3 - 5xy + 2x ) ( -

Trần Huỳnh Khả My
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Hồ Ngọc Mạnh
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Hồ Ngọc Mạnh
3 tháng 12 2021 lúc 19:40

Giúp với mọi người

 

 

Nguyễn Lê Phước Thịnh
3 tháng 12 2021 lúc 20:44

c: \(=3x^3-4x^2+6x^2-8x=3x^2+2x^2-8x\)

Võ Lan Nhi
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Phương Trâm
23 tháng 11 2017 lúc 15:10

a) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)

\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)

\(=\dfrac{\left(2x+y\right)\left(2x+y\right)-8yx+\left(2x-y\right)\left(2x-y\right)}{x\left(2x+y\right)\left(2x-y\right)}\)

\(=\dfrac{8x^2-8xy+2y^2}{x\left(2x+y\right)\left(2x-y\right)}\)

\(=\dfrac{2\left(4x^2-4xy+y^2\right)}{x\left(2x+y\right)\left(2x-y\right)}\)

\(=\dfrac{2\left(2x-y\right)^2}{x\left(2x+y\right)\left(2x-y\right)}\)

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

b) \(\dfrac{1}{x^2+3x+2}+\dfrac{2x}{x^2+4x+3}+\dfrac{1}{x^2+5x+6}\)

\(=\dfrac{1}{x^2+x+2x+2}+\dfrac{2x}{x^2+x+3x+3}+\dfrac{1}{x^2+2x+3x+6}\)

\(=\dfrac{1}{x\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{x\left(x+1\right)+3\left(x+1\right)}+\dfrac{1}{x\left(x+2\right)+2\left(x+2\right)}\)

\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{x+3+2x\left(x+2\right)+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

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

\(=\dfrac{2x^2+6x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{2\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{2\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)

\(=\dfrac{2}{x+3}\)