a: =>y-3-(48-48:6-3)=0

=>y-3-48+8+3=0

=>y-40=0

hay y=40

b: =>15x13:(29/3-y)=39

=>29/3-y=5

hay y=14/3

\(a,\)\(y-\frac{6}{2}-\left(48-24\times\frac{2}{6}-3\right)=0\)

\(\Rightarrow y-3-\left(48-8-3\right)=0\)

\(\Rightarrow y-3-48+8+3=0\)

\(\Rightarrow y-40=0\)

\(\Rightarrow y=40\)

\(\frac{3}{4\times7}+\frac{1}{7\times8}+\frac{5}{8\times13}+\frac{2}{13\times15}+\frac{9}{15\times24}\)

=  \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{24}\)

= \(\frac{1}{4}-\frac{1}{24}\)

=  \(\frac{6}{24}-\frac{1}{24}\)

= \(\frac{5}{24}\)

\(\frac{3}{4.7}+\frac{1}{1.8}+\frac{5}{8.13}+\frac{2}{13.15}+\frac{9}{15.24}\)

Đặt A = ( 3 + 1 + 5 + 2 + 9 ) . \(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{24}\)

      A = 20 . \(\frac{1}{4}-\frac{1}{24}\)

      A = 20 . \(\frac{6}{24}-\frac{1}{24}\)

      A  = 20 . \(\frac{5}{24}\)

      A  = \(\frac{100}{24}\)

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# DanLinh

Bạn Dan Linh ơi dãy số: \(\frac{3}{4\times7}+\frac{1}{7\times8}+\frac{5}{8\times13}+\frac{2}{13\times15}+\frac{9}{15\times24}\)

Không có chung ( 3 + 1 + 5 + 2 + 9 ) mà bạn lại đặt ra là sao vậy

Ta có :

\(\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}\)

\(=\)\(\frac{2\left(x+1\right)}{4}=\frac{3\left(y+3\right)}{12}=\frac{4\left(z+5\right)}{24}\)

Theo tính chất của dãy tỉ số bằng nhau,ta có :

\(\frac{2\left(x+1\right)}{4}=\frac{3\left(y+3\right)}{12}=\frac{4\left(z+5\right)}{24}\)\(=\frac{\left(2x+3y+4z\right)+\left(2+3+5\right)}{4+12+24}\)\(=\)\(\frac{9+10}{40}\)\(=\frac{19}{40}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{19}{40}\\y=\frac{19}{40}\\z=\frac{19}{40}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{19}{40}\cdot2\\y=\frac{19}{40}\cdot4\\z=\frac{19}{40}.6\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0,95\\y=1,9\\z=2,85\end{cases}}\)

Vậy ...

P/s : sai thì thôi =.=

a) \(y-6:2-\left(48-24.2:6-3\right)=0.\)

\(\Rightarrow y-3-\left(48-8-3\right)=0\)

\(\Rightarrow y-3-37=0\)

\(\Rightarrow y=0+37+3=40\)

Vậy y = 40

b) \(\left(7.13+8.13\right):\left(9\frac{2}{3}-y\right)=39\)

\(\Rightarrow195:\left(\frac{29}{3}-y\right)=39\)

\(\Rightarrow\frac{29}{3}-y=195:39=5\)

\(\Rightarrow y=\frac{29}{3}-5=\frac{14}{3}\)

Vậy y=\(\frac{14}{3}\)

a) y - 6 : 2 - ( 48 - 24 x 2 : 6 - 3 ) = 0

y - 6 : 2 - ( 48 - 48 : 6 - 3 ) = 0

y - 6 : 2 - ( 48 - 8 - 3 ) = 0

y - 6 : 2 - 37 = 0

y - 3 = 0 + 37

y - 3 = 37

y = 37 + 3

y = 40

b) ( 7 x 13 + 8 x 13 ) : (  \(9\frac{2}{3}\) - y ) = 39

( 15 x 13 ) : ( \(9\frac{2}{3}\) - y ) = 39

195 : ( \(9\frac{2}{3}\) - y ) = 39

\(9\frac{2}{3}\) - y = 195 : 39

\(9\frac{2}{3}-y=5\)

\(y=9\frac{2}{3}-5\)

\(y=\frac{14}{3}\)

 a,  y - 6 : 2 - ( 48 - 24 x 2 : 6 - 3 ) = 0 

     y - 3 - 37 = 0 

     y - 3 = 0 + 37 

     y - 3 = 37 

     y = 37 + 3

     y = 40 

b,  ( 7 x 13 + 8 x 13 ) : ( \(9\frac{2}{3}\) - y ) = 39

     195 : ( \(\frac{29}{3}\)- y ) = 39 

     \(\frac{29}{3}\)- y = 195 : 39 

   \(\frac{29}{3}\)- y = 5

  y = \(\frac{29}{3}\)- 5 

y = \(\frac{14}{3}\)

\(a)y-6:2-\left(48-24x2:6-3\right)=0\)\(0\)

    \(y-3-\left(48-48:6-3\right)=0\)

    \(y-3-\left(48-8-3\right)=0\)

    \(y-3-\left(40-3\right)=0\)

    \(y-3-37=0\)

    \(y-\left(3+37\right)=0\)

    \(y-40=0\)

    \(y=0+40\)

    \(y=40\)

\(b)\left(7x13+8x13\right):\left(9\frac{2}{3}-y\right)=39\)

      \(\left(91+104\right):\left(\frac{29}{3}-y\right)=39\)

        \(195:\left(\frac{29}{3}-y\right)=39\)

         \(\frac{29}{3}-y=195:39\)

          \(\frac{29}{3}-y=5\)

            \(y=\frac{29}{3}-5\)

            \(y=\frac{14}{3}\)