Tìm y:
(7x13+8x13):(\(9\frac{2}{3}\)-y)=39
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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
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
\(9\frac{2}{3}-y=5\)
\(y=9\frac{2}{3}-5\)
\(y=\frac{14}{3}\)
\(a)y-6:2-\left(48-24x2:6-3\right)=0\)\(0\)
\(y-3-\left(48-48:6-3\right)=0\)
\(y-3-\left(48-8-3\right)=0\)
\(y-3-\left(40-3\right)=0\)
\(y-3-37=0\)
\(y-\left(3+37\right)=0\)
\(y-40=0\)
\(y=0+40\)
\(y=40\)
\(b)\left(7x13+8x13\right):\left(9\frac{2}{3}-y\right)=39\)
\(\left(91+104\right):\left(\frac{29}{3}-y\right)=39\)
\(195:\left(\frac{29}{3}-y\right)=39\)
\(\frac{29}{3}-y=195:39\)
\(\frac{29}{3}-y=5\)
\(y=\frac{29}{3}-5\)
\(y=\frac{14}{3}\)