y= 13,5

Trả lời:

\(y\times\frac{15}{2}-\frac{1}{3}\times\left(\frac{1}{4}+y\right)=96\frac{2}{3}\)

\(\Leftrightarrow y\times\frac{15}{2}-\frac{1}{12}-\frac{1}{3}\times y=\frac{290}{3}\)

\(\Leftrightarrow y\times\left(\frac{15}{2}-\frac{1}{3}\right)=\frac{387}{4}\)

\(\Leftrightarrow y\times\frac{43}{6}=\frac{387}{4}\)

\(\Leftrightarrow y=\frac{27}{2}\)

Vậy \(y=\frac{27}{2}\)

1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)

   \(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)

 ( . là nhân nha) 

    \(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)

   \(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)

\(x=\frac{113}{8}\)

( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)

   \(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{49}{81}=\frac{56}{81}\)

\(4y=\frac{7}{81}\)

y      =  7/81:4

y       = 7/324

\(a,\)\(71+65\times4=\frac{x+140}{x}+260\)

\(\Rightarrow71+260=\frac{x-140}{x}+260\)

\(\Rightarrow71=\frac{x-140}{x}\)

\(\Rightarrow71x=x-140\)

\(\Rightarrow71x-x=-140\)

\(\Rightarrow70x=-140\)

\(\Rightarrow x=-2\)

\(b,\)\(y\times\frac{15}{2}-\frac{1}{3}\times\left(\frac{1}{4}+y\right)=90\frac{2}{3}\)

\(\Rightarrow\frac{15y}{2}-\frac{1}{12}-\frac{y}{3}=\frac{272}{3}\)

\(\Rightarrow\frac{90y}{12}-\frac{1}{12}-\frac{4y}{12}=\frac{1088}{12}\)

\(\Rightarrow90y-1-4y=1088\)

\(\Rightarrow86y=1089\)

\(\Rightarrow y=\frac{1089}{86}\)

\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé

\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)

Vậy \(A=\frac{1}{20}\)

\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)

\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)

Vậy \(B=1004\)

DẤU CHẤM LÀ DẤU NHÂN

a, 

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)

b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)

a ) 1 + 2 + 3 + 4 + ... + x = 1275 ( có x số tự nhiên )

( x + 1 ) . x : 2 = 1275

( x + 1 ) . x = 1275 x 2

( x + 1 ) . x = 2550

( x + 1 ) . x = 50 . 51 

Mà x , x + 1 là hai số tự nhiên liên tiếp => x = 50

Vậy x = 50

1+2+3+4+...+x=1275

\(\frac{x.\left(x+1\right)}{2}=1275\)

x(x+1)=1275x2=2550

x(x+1)=50.51

x=50

\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}...\frac{2018}{2016}.\frac{2019}{2017}\\ =\frac{3.4.5...2018.2019}{1.2.3...2016.2017}\\ =\frac{2018.2019}{2}=1009.2019\)