1 + 1/3 + 1/6 + 1/10 +..............+1/x.(x+1):2 = 1 và 2003/2005 ( 1/3 ; 1/5 ; ....... / 1/x.(x+1):2 là phân số nha còn 1 và 2003/2005 là hỗn số )
ai giúp mik vs, mik giải đc gần kết thì ko tính ra đc nên nhờ các bạn
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Phân số \(\dfrac{1}{4}\) ở trên là mình viết nhầm nhé. thực ra đề bài không có phân số này 😅
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
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2005}\)
\(\Leftrightarrow x+1=2005\)
\(\Leftrightarrow x=2004\)
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
ai tích mk tích lại cho
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
ai tích mk tích lại cho
( 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 ~
lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooolllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)
\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)
\(\frac{1}{x+1}=\frac{1}{2003}\)
\(\Rightarrow x+1=2003\)
\(x=2002\)
Vậy x = 2002
a, 1 - 2 - 3 + 4 + 5 + 6 - 7 + ... + 1996 + 1997 - 1998 - 1999 + 2000 + 2001
= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ... +( 1993 - 1994 - 1995 + 1996 ) + ( 1997 - 1998 - 1999 + 2000 ) + 2001
= 0 + 0 + ... + 0 + 0 + 2001
= 2001
b, 1 - 3 + 5 -7 + ...+ 2001 - 2003 + 2005
= ( 1 -3 ) + ( 5 - 7 ) + ... + ( 2001 - 2003 ) + 2005
= -2 + (-2 )+ ... + (-2) + 2005
Có 501 số ( - 2 )
= ( - 2 ) . 501 + 2005
= -1002 + 2005
= 1003
P/s : Tham khảo ( Chép đầu bài thôi cũng sai )