rắn <lỏng <khí

nhớ like nhé

cái thớt 

**** cho mình nha

thiếu j              

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

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

4(x-2)-5(4-x)=6

4.x-4.2-5.4-5.x=6

4x-8-20-5x=6

9x=6+8-20

9x=-6

x=-6:9

x=-2/3

3002+433=3435

1576+1503=3070

7566+1452=9018

8520+3240=11760

9826+6059=10435

đúng 1000% trên 100%