x + 9x + 7x + 5x = 2244

1x + 9x + 7x + 5x = 2244

x(1+9+7+5) = 2244

x.22 = 2244

=> x = 102

x + 9x + 7x + 5x = 2244

<=> 1x + 9x + 7x + 5x = 2244

x ( 1 + 9 + 7 + 5 ) = 2244

22x = 2244

x = 2244 : 22

x = 1122

x + 9x + 7x + 5x = 2244

<=> 22x = 2244

<=> x = 2244 : 22 

=> x = 102

66666666666666666666666666666666

a: \(=\dfrac{2x^5-2x^3-3x^3+3x+x^2-1}{x^2-1}\)

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

1*x+9*x+7*x+5*x =8844

(1+9+7+5)*x=8844

22*x=8844

x=8844/22

x=402

x=402 nha k nha

a) x² + 5x + 6

= x² + 2x + 3x + 6

= (x² + 2x) + (3x + 6)

= x(x + 2) + 3 (x + 2)

= (x + 2) (x + 3)

b) x² + 6x + 8

= x² + 2x + 4x + 8

= (x² + 2x) + (4x + 8)

= x(x + 2) + 4(x + 2)

= (x + 2)(x + 4)

c) x² - 5x - 14

= x² + 2x - 7x - 14

= (x² + 2x) - (7x + 14)

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

= (x + 2)(x - 7)

d) x² - 9x + 18

= x² - 3x - 6x + 18

= (x² - 3x) - (6x + 18)

= x(x - 3) - 6 (x - 3)

= (x - 3)(x - 6)

e) x² - 7x + 12

= x² -3x - 4x + 12

= (x² - 3x) - (4x + 12)

= x(x - 3) - 4(x - 3)

= (x - 3)(x - 4)

f) 3x² + 9x - 30

= 3(x² + 3x - 10)

= 3\(\left[\left(x^2+5x-2x-10\right)\right]\)

= 3\(\left[\left(x^2+5x\right)-\left(2x-10\right)\right]\)

= 3\(\left[x\left(x+5\right)-2\left(x+5\right)\right]\)

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

 Chuc ban hoc tot

a) \(x^2+5x+6=\left(x+2\right)\left(x+3\right)\)

b) \(x^2+6x+8=\left(x+2\right)\left(x+4\right)\)

c) \(x^2-5x-14=\left(x-7\right)\left(x+2\right)\)

d) \(x^2-9x+18=\left(x-3\right)\left(x-6\right)\)

e) \(x^2-7x+12=\left(x-3\right)\left(x-4\right)\)

f) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x+5\right)\left(x-2\right)\)

Đề yêu cầu gì em?

\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)

\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)

\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)

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

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

b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)

\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)

thiếu đề bài nhé bạn