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=(1+...2005)x(125x1001x127-127x1001x125)
=(1+...2005)x0(cả hai vế giống nhau nên trừ đi thì =0)
=0
\(a,\)\(y-\frac{6}{2}-\left(48-24\times\frac{2}{6}-3\right)=0\)
\(\Rightarrow y-3-\left(48-8-3\right)=0\)
\(\Rightarrow y-3-48+8+3=0\)
\(\Rightarrow y-40=0\)
\(\Rightarrow y=40\)
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
( 1+3+5+7+…+2003+2005) x (125 125 x 127 – 127 127 x 125)
= ( 1+3+5+7+…+2003+2005) x (125 x 1001 x 127 – 127 x 1001x 125)
= ( 1+3+5+7+…+2003+2005) x 0 = 0
bài 1 : 0
bài 2 : ( 8.78 + 1.25 ) x y= 26.3
10.03 x y = 26.3
y = 26.3 : 10.03
y = 2630/1003
bài 3 :diện tích HTG ABC là :
12 x 18 : 2 = 108 ( cm2 )
quên rùi
( 1 + 3 + 5 + 7 +... + 2003 + 2005 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 125 x 1001 x 127 - 127 x 1001 x 125 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 0
= 0
~ Thiên Mã ~
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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
Chúc bạn học tốt
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 \(y=40\)