\(a,\dfrac{17}{18}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{9}\\ b,\dfrac{9}{12}=\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{3}\)

Bổ xung ý 2

\(\dfrac{1}{x}+\dfrac{y}{3}=\dfrac{5}{6}\\ \Rightarrow\dfrac{1}{x}=\dfrac{5}{6}-\dfrac{y}{3}\\ \Rightarrow\dfrac{1}{x}=\dfrac{5-2y}{6}\\ \Rightarrow x\cdot\left(5-2y\right)=6\)

`=>x;5-2y in Ư(6)={+-1;+-3;+-2;+-6}`

mà `5-2y` là số lẻ 

nên `5-2y in {+-1;+-3}`

Ta có bảng sau :

Vậy `x;y in {(1;2);(3;1)}`

ở 11B*B là +1/11^99 nha 

\(\dfrac{5^{x+1}}{125}=\dfrac{1}{25^{x-2}}\\ \dfrac{5^{x+1}}{125}=\dfrac{1}{5^{2x-4}}\\ 5^{x+1}\cdot5^{2x-4}=125\\ 5^{x+1+2x-4}=5^3\\ 5^{\left(x+2x\right)+\left(1-4\right)}=5^3\\ 5^{3x-3}=5^3\\ 3x-3=3\\ 3x=6\\ x=2\)

\(B=\dfrac{1}{11}+\dfrac{1}{11^2}+\dfrac{1}{11^3}+...+\dfrac{1}{11^{99}}+\dfrac{1}{11^{100}}\\ 11B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{11^{98}}+\dfrac{1}{11^{99}}\\ 11B-B=1+\dfrac{1}{11}+\dfrac{1}{11^2}+...+\dfrac{1}{1^{99}0}-\dfrac{1}{11}-\dfrac{1}{11^2}-\dfrac{1}{11^3}-...-\dfrac{1}{11^{100}}\\ 10B=1-\dfrac{1}{11^{99}}\\ B=\dfrac{1-\dfrac{1}{11^{99}}}{10}\)

có : `1-1/(11^99)<1`

\(\Rightarrow\dfrac{1-\dfrac{1}{11^{99}}}{10}< \dfrac{1}{10}\)

hay `B<1/10`

bạn xem lại nhé!

thấy -21:7=-3 , mẫu p/s 2 còn 1

24:(-3)=-8 , tử phân số 2 còn -1