mày vớ vẩn

1+1=2 các bạn

B=(1/4+1/5+1/6+...+1/9)+(1/10+1/11+...+1/19)

1/4+1/5+1/6+1/7+1/8+1/9>1/9+1/9+1/9+1/9+1/9+1/9=6/9>1/2

1/10+1/11+...+1/19>1/19+1/19+...+1/19=10/19>1/2

                                       10 số          

=>B>1/2+1/2=1

Vậy ta có ĐPCM

giản dị ,trung thực,hòa đồng ,hiếu học,đoàn kết với bạn bè

\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=ab^2-ac^2+bc^2-a^2b+a^2c-b^2c\)

\(=ab\left(b-a\right)+c^2\left(b-a\right)+c\left(a^2-b^2\right)\)

\(=\left(b-a\right)\left(ab+c^2\right)-c\left(b^2-a^2\right)\)

\(=\left(b-a\right)\left(ab+c^2\right)-c\left(b-a\right)\left(b+a\right)\)

\(=\left(b-a\right)\left[ab+c^2-c\left(b+a\right)\right]\)

\(=\left(b-a\right)\left(ab+c^2-bc-ac\right)\)

\(=\left(b-a\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)

\(=\left(b-a\right)\left(a-c\right)\left(b-c\right)\)

\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left(a^2-b^2+b^2-c^2\right)+c\left(a^2-b^2\right)\)

\(=a\left(b^2-c^2\right)-b\left(a^2-b^2\right)-b\left(b^2-c^2\right)+c\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b^2-c^2\right)+\left(c-b\right)\left(a^2-b^2\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left[b+c-\left(a+b\right)\right]\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

18% của 100  là :   

    18 x 100 : 100 = 18

ai mình mình lại