Ta có:

Xét: Ba = Bố 

Mà Ba n = Bố n = Bốn 

~ Hok T ~

vào cái này nha https://vn.lovemeter.me/match/gWWf

vì\(9!-2!=362878\)

Mà \(7!=5040\)

Nên \(9!-2!\ne7!\)

mk cũng bị hết lượt ( T_T )

Bài 1:

\(a,3x^2-6x+9x^3=3x\left(x-2+3x^2\right)=3x\left[\left(3x^2+3x\right)-\left(2x+2\right)\right]=3x\left(x+1\right)\left(3x-2\right)\\ b,2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(x+y\right)\left(2-x\right)\\ c,x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)

\(d,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(y-x\right)\left(x^2-9\right)=\left(y-x\right)\left(x-3\right)\left(x+3\right)\\ e,x^2-25+y^2+2xy=\left(x+y\right)^2-5^2=\left(x+y-5\right)\left(x+y+5\right)\\ f,\left(x^2+1\right)^2-4x^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)

\(g,x^2\left(x-1\right)+16\left(1-x\right)=\left(x-1\right)\left(x^2-16\right)=\left(x-1\right)\left(x-4\right)\left(x+4\right)\\ h,5x\left(x-2y\right)+2\left(2y-x\right)^2=5x\left(x-2y\right)+2\left(x-2y\right)^2=\left(x-2y\right)\left[5x+2\left(x-2y\right)\right]=\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)

a: \(=3x\left(3x^2+x-2\right)\)

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

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

😨 😓 😓 tui là con gái -> hok hành / hết

bố mẹ ở nhà còng lưng nuôi ăn hok , yiu vs đương cái j , yiu vs đương cái j ???

- tầm này còn quá sớn -

#girl_2k5_#

làm ny ó bn 👌 ❤️ ❤️

1+1=2 

hok tốt

.... kb òi k ik

Trả lời :

1 + 1

= 2

Mk kb ròi :))

Hok tốt !

#Minh#

bn có thi olympic à, kb vs mk nha

uk, có nhưng mik hết lượt kb rùi

kinh chưa

Mình thử nha :33

ĐKXĐ : \(x\ne-3,x\ne-26,x\ne-6,x\ne1\)

Ta có :

\(A=\left[\frac{3}{2}-\left(\frac{x^4\left(x^2+1\right)-x^4-1}{x^2+1}\right)\cdot\frac{x^3-4x^2+\left(x-4\right)}{x^6\left(x+6\right)-\left(x+6\right)}\right]:\frac{\left(x+3\right)\left(x+26\right)}{3\left(x-2\right)\left(x+6\right)}\)

\(=\left[\frac{3}{2}-\left(\frac{x^6-1}{x^2+1}\right)\cdot\frac{\left(x-4\right)\left(x^2+1\right)}{\left(x+6\right)\left(x^6-1\right)}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)

\(=\left[\frac{3}{2}-\frac{x-4}{x+6}\right]\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)

\(=\frac{x+26}{2\left(x+6\right)}\cdot\frac{3\left(x-2\right)\left(x+6\right)}{\left(x+3\right)\left(x+26\right)}\)

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

Vậy : \(A=\frac{3\left(x-2\right)}{2\left(x+3\right)}\left(x\ne-3,x\ne-26,x\ne-6,x\ne1\right)\)

ngầu rồi