6B, 7A, 8D

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

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

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

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

1.(2x+3).(x-5)+2x(3-x)+x-10

=2x^2 -10x+3x-15+6x-2x^2+x-10

=2x-25

2.(-x-2)³+(2x-4).(x²+2x+4)-x².(x-6)

=-x^3+6x^2-12x-8+2x^3+4x^2+8x-4x^2+8x-16-x^3+6x^2

bổ sung câu 2 tiếp là

=12x^2 + 4x-24

a. 5x + 3(x² - x - 1)

= 5x + 3x² - 3x - 3

= 3x² + 5x - 3x - 3

= 3x² + 2x - 3

b. (5 - x)(5 + x) - (2x - 1)²

25 - x² - (4x² - 4x + 1)

= 25 - x² - 4x² + 4x - 1

= 25 - 1 - x² - 4x² + 4x 

= 24 - 5x² + 4x

a) \(5x+3\left(x^2-x-1\right)=5x+3x^2-3x-3=3x^2+2x-3\)

b) \(\left(5-x\right)\left(5+x\right)-\left(2x-1\right)^2=25-x^2-4x^2+4x-1=-5x^2+4x+24\)

a) \(=6x^3+8x^2+2x-6x^3=8x^2+2x\)

b) \(=\left[3xy\left(xy+2xy^2-4\right)\right]:3xy=xy+2xy^2-4\)

c) \(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x+2}-\dfrac{5}{x-2}=\dfrac{10x+3\left(x-2\right)-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)

a, \(=6x^3+12x^2+2x-6x^3\\=12x^2+2x\)

b,

\(=xy+2xy^2-4\)

c,

\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)

\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)

Dễ

 Thế

Mà

Cũnhoir

Dc

Ạ

Chịu

Chắc

Phải

Ngu 

Lamqs

Mới

Hỏi

Câu

Này

\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)

1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)

2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)

3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)