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NL
17 tháng 7 2020
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
17 tháng 7 2020
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 \(y=40\)
TT
0
TC
4
H9
HT.Phong (9A5)
CTVHS
16 tháng 10 2023
\(\dfrac{1}{2001\times2003}+\dfrac{1}{2003\times2005}+...+\dfrac{1}{2011\times2013}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{2001\times2003}+\dfrac{2}{2003\times2005}+...+\dfrac{2}{2011\times2013}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{4}{1342671}\)
\(=\dfrac{2}{1342671}\)
T
16 tháng 10 2023
\(\dfrac{1}{2001\times2003}+\dfrac{1}{2003\times2005}+\dfrac{1}{2005\times2007}+...+\dfrac{1}{2011\times2013}\) (sửa đề)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{2001\times2003}+\dfrac{2}{2003\times2005}+\dfrac{2}{2005\times2007}+...+\dfrac{2}{2011\times2013}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+\dfrac{1}{2005}-\dfrac{1}{2007}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)\)
\(=\dfrac{1}{2}\times\dfrac{4}{1342671}\)
\(=\dfrac{2}{1342671}\)
DV
0
TA
1
U
10 tháng 8 2018
\(\frac{1}{2001\cdot2003}+\frac{1}{2003\cdot2004}+...+\frac{1}{2011\cdot2013}\)
\(=\frac{1}{2}\left(\frac{2}{2001\cdot2003}+\frac{2}{2003\cdot2005}+...+\frac{2}{2011\cdot2013}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2001}-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2005}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2001}-\frac{1}{2013}\right)\)
tự tính tiếp
1/3 + 1/6 + 1/10 + ... + 2/x(x+1) = 2003/2005
2 × ( 1/6 + 1/12 + 1/20 + ... + 1/x(x+1) = 2003/2005
1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x+1) = 2003/2005 : 2
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 2003/2005 × 1/2
1/2 - 1/x+1 = 2003/4010
1/x+1 = 1/2 - 2003/4010
1/x+1 = 2005/4010 - 2003/4010
1/x+1 = 1/2005
=> x+1 = 2005
=> x = 2004
Vậy x = 2004
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2005}\)
\(\Leftrightarrow x+1=2005\)
\(\Leftrightarrow x=2004\)