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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 : \(x\ne\pm3\)

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

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

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

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

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

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

23 tháng 7 2023

\(a,\dfrac{x^2-9}{x-2}:\dfrac{x-3}{x}\\ =\dfrac{\left(x-3\right)\left(x+3\right)}{x-2}\times\dfrac{x}{x-3}\\ =\dfrac{x\left(x+3\right)}{\left(x-2\right)}\)

\(b,\dfrac{x}{z^2}.\dfrac{xz}{y^3}:\dfrac{x^3}{yz}\\ =\dfrac{x}{z^2}.\dfrac{xz}{y^3}.\dfrac{yz}{x^3}=\dfrac{x^2yz^2}{z^2y^3x^3}=\dfrac{1}{xy^2}\)

\(c,\dfrac{2}{x}-\dfrac{2}{x}:\dfrac{1}{x}+\dfrac{4}{x}.\dfrac{x^2}{2}\\ =\dfrac{2}{x}-\dfrac{2}{x}\times\dfrac{x}{1}+\dfrac{4x^2}{2x}\\ =\dfrac{2}{x}-\dfrac{2}{1}+2x\\ =\dfrac{2-2x+2x^2}{x}\)

23 tháng 7 2023

a) \(\dfrac{x^2-9}{x-2}:\dfrac{x-3}{x}\)

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

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

b) \(\dfrac{x}{z^2}\cdot\dfrac{xz}{y^3}:\dfrac{x^3}{yz}\)

\(=\dfrac{x}{z^2}\cdot\dfrac{xz}{y^3}\cdot\dfrac{yz}{x^3}\)

\(=\dfrac{1}{xy^2}\)

c) \(\dfrac{2}{x}-\dfrac{2}{x}:\dfrac{1}{x}+\dfrac{4}{x}\cdot\dfrac{x^2}{2}\)

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

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

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

\(=\dfrac{6}{x}-2\)

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

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

23 tháng 7 2023

\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)

\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)

\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)

\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)

\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)

17 tháng 10 2023

a) \(x\left(x^2+4x+5\right)-x^2\left(x+4\right)\)

\(=x^3+4x^2+5x-x^3-4x^2\)

\(=5x\)

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

\(=x^2-4x+4+3x-3-x^2+x\)

\(=1\)

c) \(\left(x+2\right)^3-x\left(x^2+6x+12\right)\)

\(=x^3+6x^2+12x+8-x^3-6x^2-12x\)

\(=8\)

 

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:

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