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

2: 

a: =>-2x=10

=>x=-5

b: =>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

6 tháng 8 2020

cau hoi nay de lam ma

6 tháng 8 2020

\(\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{x^2-9}\)

ĐKXĐ : \(x\ne\pm3\)

\(=\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}+\frac{5x-15}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{4x+12+5x-15-13+9x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\frac{9x^2+9x-16}{\left(x+3\right)\left(x-3\right)}=\frac{9x^2+9x-16}{x^2-9}\)

11 tháng 12 2020

a) ( 3 + x )2 = x2 + 6x + 9

b) ( 5 - x )3 = 125 - 75x + 45x2 - x3

c) ( 2x - 1 )( x2 - x + 3 ) = 2x3 - 2x2 + 6x - x2 + x - 3 = 2x3 - 3x2 + 7x - 3

d) \(\frac{9}{x^2+3x}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}+\frac{x-3}{x}\)

\(=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

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

11 tháng 12 2020

a, \(\left(3+x\right)^2=9+6x+x^2\)

b, \(\left(5-x\right)^3=125-75x+15x^2-x^3\)

c, \(\left(2x-1\right)\left(x^2-x+3\right)=2x^3-2x^2+6x-x^2+x-3=2x^3-3x^2+7x-3\)

d, \(\frac{9}{x^2+3x}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}-\frac{3-x}{x}=\frac{9}{x\left(x+3\right)}+\frac{\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

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

12 tháng 8 2015

1)(x+1)(x+2)(x+3)=x3+6x2+11x+6

 

12 tháng 8 2015

2)a)3x(x-5)-(x-1)(2+3x)=30

<=>3x2-15x-3x2+x+2=30

<=>-14x+2=30

<=>-14x=30-2

<=>-14x=28

<=>x=-2

b)(x+2)(x+3)-(x-2)(x+5)=0

<=>x2+5x+6-x2-3x+10=0

<=>2x+16=0

<=>2x=-16

<=>x=-8

c)(3x+2)(2x+9)-(x+2)(6x+1)=9

<=>6x2+31x+18-6x2-13x-2=9

<=>18x+16=9

<=>18x=9-16

<=>18x=-7

<=>x=-7/18

21 tháng 12 2021

a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)

\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)

\(=7x-16\)

21 tháng 12 2021

b, = x2 - 16 - ( x3 - 33 ) : ( x - 3 )

= x2 - 16 - \([\) ( x - 3 ) ( x2 + 3x + 9 ) \(]\) : ( x - 3 )

= x2 - 16 - ( x2 + 3x + 9 )

= x2 - 16 - x2 - 3x - 9

= -25 - 3x

9 tháng 9 2017

a,       x- 9 - x2 - 3x + 10 = 1 - 3x

8 tháng 8 2020

\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)

\(=x^3-8-\left(x^3-3x^2+3x-1\right)+7\)

\(=x^3-8+7-x^3+3x^2-3x+1\)

\(=\left(x^3-x^3\right)+\left(7+1-8\right)+3x^2-3x\)

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

8 tháng 8 2020

\(x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x\left(2+x\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x\left(4-x^2\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=4x-x^3+\left(x^3+9\right)\)

\(=4x-\left(x^3-x^3\right)+9\)

\(=4x+9\)

6 tháng 8 2018

\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)

\(=4x\left(x^2-9\right)-x^3+27\)

\(=4x^3-36x-x^3+27\)

\(=3x^3-36x+27\)

6 tháng 8 2018

\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)

\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)

\(=\left(x+6\right).0\)

\(=0\)