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24 tháng 11 2019

a) \(\frac{x^2+2x-3}{x^2+x-6}\)

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

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

\(=\frac{\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x-1}{x-2}\)

b) \(\frac{2x^2+2x+2}{x^2+6x-7}\)

\(=\frac{2\left(x^2+x+1\right)}{x^2+7x-x-7}\)

\(cogidosaisai?\)

10 tháng 10 2021

b: Ta có: \(\left(x-3\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+9\left(x+2\right)^2\)

\(=x^3-9x^2+27x-27-x^3-8+9x^2+36x+36\)

\(=53x+1\)

a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)

\(=6x^2-2x-6x^2-2x+18x+6\)

=14x+6

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

\(=4x^2-12x+9-4x^2+1+6x-9\)

\(=-6x+1\)

c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)

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

=1

4 tháng 9 2021

a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)

b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)

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

19 tháng 7 2016

\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)

\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)

 

19 tháng 7 2016

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)

31 tháng 12 2019

a) (2x - 1)(3x + 5) - 2(-4x + 1)2 = 6x2 + 10x - 3x - 5 - 2(16x2 - 8x + 1) = 6x2 - 3x - 5 - 32x2 + 16x - 2 = -26x2 + 13x - 7

b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)

c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)

\(\frac{2x-9}{x^2-2x-3x+6}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}-\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\)

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

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

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

\(\frac{\left(x+1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\frac{x+1}{x-3}\)

d) (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)

= (x - 1 - x - 1)[(x - 1)2 + (x - 1)(x + 1) + (x + 1)2] + 6(x2 - 1)

= -2(x2 - 2x + 1  + x2 - 1 + x2 + 2x + 1) + 6x2 - 6

= -2(3x2 + 1) + 6x2 - 6

= -6x2 - 2 + 6x2  - 6

= -8

e) (2x + 7)2 - (4x + 14)(2x - 8) + (8 - 2x)2

= (2x + 7)2 - 2(2x + 7)(2x - 8) + (2x - 8)2

= (2x + 7 - 2x + 8)2

= 152 = 225

27 tháng 2 2020

1)\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3x}{\left(2x+6\right)x}-\frac{x-6}{2x^2+6x}\\ =\frac{3x}{2x^2+6x}-\frac{x-6}{2x^2+6x}=\frac{3x-\left(x-6\right)}{2x^2+6x}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)

20 tháng 1 2019

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

20 tháng 1 2019

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19