Lời giải:
$60\text{%}\times y+\frac{3}{5}\times y+y=19,8$

$\frac{3}{5}\times y+3,5\times y+y=19,8$

$y\times (\frac{3}{5}+\frac{3}{5}+1)=19,8$

$y\times 2,2=19,8$

$y=19,8:2,2=9$

\(\Rightarrow y\times\left(2+\dfrac{1}{5}\right)=\dfrac{8}{5}\\ \Rightarrow y\times\dfrac{11}{5}=\dfrac{8}{5}\\ \Rightarrow y=\dfrac{8}{5}:\dfrac{11}{5}=\dfrac{8}{5}\times\dfrac{5}{11}=\dfrac{8}{11}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{7}{8}=\dfrac{1}{8}\)

`= 1/2 xx 2/3 xx...xx 7/8`

`= 1/8`

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{6}{7}\times\dfrac{7}{8}\\ =\dfrac{1}{8}\)

\(1152-96\cdot y=\left(96\cdot y-96\cdot y\right)\cdot96\cdot y\)(dấu nhân viết là ".")

\(1152-96\cdot y=1\cdot96\cdot y\)

\(1152=96\cdot y+96\cdot y\)

\(1152=192\cdot y\)

\(y=\frac{1152}{192}\)

\(y=6\)

(1,5+1).y=2,5+7,5

2,5.y=10

y=10:2,5

y=4

1,5 x y + y = 2,5 + 3 ✖︎ 2,5

 1,5 x y + y =  2,5 + 7,5

1,5 x y + y = 10

y x ( 1,5 + 1) = 10

y x 2,5 = 10

y = 10 : 2,5

y = 4

1,5 x y + y = 2,5 + 3 x 2,5

1,5 x y + y =  2,5 + 7,5.

y x ( 1,5 + 1) = 10

\(=>y=\dfrac{10}{1,5+1}=\dfrac{10}{2,5}=4\)

Bài 1: 

=>y-69,3=20,5

hay y=89,8

Bài 2: 

a: \(=365.4:43.5-7.7\cdot0.2\)

=8,4-1,54=6,86

b: \(=\dfrac{2}{5}:\dfrac{8-5}{10}+\dfrac{3}{5}=\dfrac{2}{5}\cdot\dfrac{10}{3}+\dfrac{3}{5}=\dfrac{20+9}{15}=\dfrac{29}{15}\)

\(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot\left(1-\dfrac{1}{15}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\)

\(B=\left(1-\dfrac{1}{3+6+10+15+...+780}\right)\)

\(B=\left(1-\dfrac{1}{\left(780-3\right)\div3+1}\right)\)

\(B=\left(1-\dfrac{1}{260}\right)\)

\(B=\dfrac{259}{260}\)

Ta có: \(\dfrac{-5}{13}\cdot\dfrac{3}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)

\(=\dfrac{5}{13}\left(-\dfrac{3}{7}+\dfrac{1}{7}\right)-\dfrac{2}{17}\cdot\dfrac{8}{13}\)

\(=\dfrac{5}{13}\cdot\dfrac{-2}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}\)

\(=\dfrac{-10}{91}-\dfrac{16}{221}\)

\(=\dfrac{-282}{1547}\)

\(\dfrac{-5}{13}.\dfrac{3}{7}-\dfrac{2}{17}.\dfrac{8}{13}+\dfrac{5}{13}.\dfrac{1}{7}\) 

\(=\dfrac{5}{13}.\dfrac{-3}{7}-\dfrac{2}{17}.\dfrac{8}{13}+\dfrac{5}{13}.\dfrac{1}{7}\)

\(=\dfrac{5}{13}.\left(\dfrac{-3}{7}+\dfrac{1}{7}\right)+\dfrac{-2}{17}.\dfrac{8}{13}\) 

\(=\dfrac{5}{13}.\dfrac{-2}{7}+\dfrac{-2}{13}.\dfrac{8}{17}\)

\(=\dfrac{-2}{13}.\dfrac{5}{7}+\dfrac{-2}{13}.\dfrac{8}{17}\) 

\(=\dfrac{-2}{13}.\left(\dfrac{5}{7}+\dfrac{8}{17}\right)\) 

\(=\dfrac{-2}{13}.\dfrac{141}{119}\) 

\(=\dfrac{-282}{1547}\)