Theo bài ra ta có : 

\(a:b:c=3:2:2\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{2}=\dfrac{c}{3}\) và \(a+b-c=99\)

Áp dụng tính chất của dãy tỉ số bằng nhau : 

\(\dfrac{a}{2}=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{a+b-c}{2+2-3}=\dfrac{99}{1}=99\)

\(+)\)\(\dfrac{a}{2}=99\Rightarrow a=99\times2=198\)

\(+)\)\(\dfrac{b}{2}=99\Rightarrow b=99\times2=198\)

\(+)\)\(\dfrac{c}{3}=99\Rightarrow c=99\times3=297\)

Vậy \(a=198;b=198;c=297\)

sao toàn xong trc toi z:))

\(\dfrac{3}{2}+\dfrac{3}{2}\)

\(=\dfrac{3+3}{2}\)

\(=\dfrac{6}{2}=3\)

\(\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{7}.\dfrac{9}{13}+\dfrac{3}{13}.\dfrac{6}{7}\)

\(=\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{9}{13}+\dfrac{3}{13}\right)\)

\(=\dfrac{6}{7}.\left(\dfrac{8+9+3}{13}\right)\)

\(=\dfrac{6}{7}.\dfrac{20}{13}\)

\(=\dfrac{6.20}{7.13}\)

\(=\dfrac{120}{91}\)

\(\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}\)

\(=1+1+1+1\)

\(=2+1+1\)

\(=3+1\)

\(=4\)