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25 tháng 12 2021

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

25 tháng 12 2021

\(\dfrac{x^3-3x^2+3x-1}{1-x+x^2y-xy}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-a\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\)

a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

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

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

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

\(=2x^2-2y\)

19 tháng 12 2021

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

20 tháng 12 2021

cho mik xin nốt mấy câu còn lại đi bạn

 

21 tháng 10 2021

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

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

16 tháng 12 2022

f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)

g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)

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

n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)

p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)

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

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

17 tháng 10 2021

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

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

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

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

31 tháng 12 2021

\(A=\dfrac{3x}{x-1}+\dfrac{2}{x+1}+\dfrac{3-3x-2x^2}{x^2-1}.\) \(\left(ĐKXĐ:x\ne1;x\ne-1\right).\)

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

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

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

 

27 tháng 12 2020

a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)

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

=2x

b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)

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

\(=\dfrac{x+1}{3x}\)

c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)

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

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

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

27 tháng 12 2020

a, \(\dfrac{2x^2-2x}{x-1}=\dfrac{2x\left(x-1\right)}{x-1}=2x\) ( đk : \(x\ne1\) )

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

c

 

=

13 tháng 8 2023

a) (3x - 2)2 - (1 + 5x)2

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

= (-2x - 3)(8x - 1)

b) (3x + 4)(3x - 4) - (5 - x)2

= (3x)2 - 42 - (25 - 10x + x2)

= 9x2 - 16 - 25 + 10x - x2

= 8x2  + 10x - 41

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

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

\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)

\(=4x+25\)

a: =9x^2-12x+4-25x^2-10x-1

=-16x^2-22x+3

b: =9x^2-16-x^2+10x-25

=8x^2+10x-41

c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)