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

\(=9x^2+6x+1-3x\left(x^2-x-2\right)\)

\(=9x^2+6x+1-3x^3+3x^2+6x\)

\(=-3x^3+12x^2+12x+1\)

a: Ta có: \(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)

\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)

\(=-16x+8\)

b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)

\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)

=27x-55

28 tháng 8 2021

\(M=\dfrac{x^2}{x^2-3x}\left(x\ne0;x\ne3\right)\\ M=\dfrac{x^2}{x\left(x-3\right)}\\ M=\dfrac{x}{x-3}\)

\(N=\dfrac{x}{x+1}+\dfrac{3x+1}{x^2-1}\left(x\ne\pm1\right)\\ N=\dfrac{x-1+3x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{4x}{\left(x+1\right)\left(x-1\right)}\)

28 tháng 8 2021

Còn tiếp:

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

8 tháng 11 2021

làm ơn giúp em với

8 tháng 11 2021

\(a,=x^2+6x+9+2x^2+5xy^2=3x^2+6x+5xy^2+9\\ b,=9x^2-12x+4-9x^2+1=-12x+5\)

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

 

=

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

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

31 tháng 8 2020

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

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

\(=\left(x^2+2\right)^2\)

31 tháng 8 2020

               Bài làm :

Ta có :

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

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

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

\(=\left(x^2+2\right)^2\)

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