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a) y - 6 : 2 - (48 - 24 x 2 : 6 - 3) = 0
y - 3 - (48 - 48 : 6 - 3) = 0
y - 3 - (48 - 8 - 3) = 0
y - 3 - 37 = 0
y - ( 3+37) = 0
y - 40 =0
y =0+40
Y =40
b) ( 7x13 - 8x13) : ( 9 2/3 -y) =39
(7+8)x13 : (29/3 - y) =39
15 x 13 : (29/3-y) =39
195 : (29/3 - y) =39
29/3 - y =195 : 39
29/3 - y = 5
y = 29/3 - 5
y = 14/3
1 )
= (1 + 3+ 5+ .....+2003+2005) \(\times\)( 125 nhân 1001 NHÂN 127 - 127 nhân 1001 nhân 125 )
= (1 + 3+ 5+ .....+2003+2005) \(\times\)0
= 0
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Trả lời:
Bài 1
\(\left(1+3+5+...+2003+2005\right)\times\left(125125\times127-127127\times125\right)\)
\(=\left\{\left(2005+1\right)\times\left[\left(2005-1\right)\div2+1\right]\div2\right\}\times\left(125\times1001\times127-127\times1001\times125\right)\)
\(=\left(2006\times1003\div2\right)\times0\)
\(=10061009\times0\)
\(=0\)
Bài 2
\(y-6\div2-\left(48-24\times2\div6-3\right)=0\)
\(y-3-\left(48-8-3\right)=0\)
\(y-3-37=0\)
\(y-40=0\)
\(y=40\)
Vậy \(y=40\)
y - 6 : 2 - ( 48 - 48 : 6 - 3 ) = 0
y - 6 : 2 - ( 48 - 8 - 3 ) = 0
y - 6 : 2 - 37 = 0
y - 3 - 37 = 0
y = 37 + 3
y = 40
\(a)y-6:2-\left(48-24x2:6-3\right)=0\)\(0\)
\(y-3-\left(48-48:6-3\right)=0\)
\(y-3-\left(48-8-3\right)=0\)
\(y-3-\left(40-3\right)=0\)
\(y-3-37=0\)
\(y-\left(3+37\right)=0\)
\(y-40=0\)
\(y=0+40\)
\(y=40\)
\(b)\left(7x13+8x13\right):\left(9\frac{2}{3}-y\right)=39\)
\(\left(91+104\right):\left(\frac{29}{3}-y\right)=39\)
\(195:\left(\frac{29}{3}-y\right)=39\)
\(\frac{29}{3}-y=195:39\)
\(\frac{29}{3}-y=5\)
\(y=\frac{29}{3}-5\)
\(y=\frac{14}{3}\)
\(a,\)\(y-\frac{6}{2}-\left(48-24\times\frac{2}{6}-3\right)=0\)
\(\Rightarrow y-3-\left(48-8-3\right)=0\)
\(\Rightarrow y-3-48+8+3=0\)
\(\Rightarrow y-40=0\)
\(\Rightarrow y=40\)