Tìm x thuộc N biết: 1/3+1/6+1/10+...+1/x(x+1):2=2001/2003
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tìm x thuộc N 1/3+1/6+1/10+......1/x(x-1)/2=2001/2003
1 tìm x thuộc N biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...................+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
tìm số tự nhiên x biết:
1/3+1/6-1/10+...+1/x(x+1):2=2001/2003
Ta có:
1/3 + 1/6 + 1/10 + ... + 1/x(x+1):2 = 2001/2003
=> 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 2001/2003
=> 2 [1/6 + 1/12 + 1/20 + ... + 1/x(x+1)] = 2001/2003
=> 2 [1/2x3 + 1/3x4 + 1/4x5 + ... + 1/x+(x+1)] = 2001/2003
=> 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1= 2001/2003 : 2
=> 1/2 - 1/x+1 = 2001/4006
=> 1/x+1 = 1/2 - 2001/4006 = 1/2003
=> x+1 = 2003 = 2002 + 1
=>x = 2002
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Tìm số tự nhiên x biết:1/3+1/6-1/10+.....+1/x(x+1):2=2001/2003
Tìm x,biết:
1/3+1/6+1/10+...+1/x(x+1):2=2001/2003
Tìm x biết :
1/3+1/6+1/10+••••+1/x+(x+1):2=2001/2003
Câu hỏi
1/3+1/6+1/10+...+1/x(x+1):2=2001/2003
=Tôi ko biết nên tôi xin hết
Tron vn
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tìm x thuộc n biết 1/3+1/6+1/10+...+2/x.(x+1)=1999/2001
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{2001}:2=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{2001}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
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\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+..+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(\frac{1}{6}+\frac{1}{12}+..+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}:\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)
\(\frac{1}{x+1}=\frac{1}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
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tìm x thuộc N biết 1/3+1/6+1/10+...+2/x.(x+1)=1999/2001
1/3 + 1/6 + 1/10 +............+1/x(x+1):2 =2001/2003. Tìm x
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2001}{2003}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{2003}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}=\frac{1}{2003}\)
=> x + 1 = 2003
=> x = 2003 - 1
=> x = 2002
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13+16+110+...+1x(x+1):2=2001200313+16+110+...+1�(�+1):2=20012003
26+212+220+...+2x(x+1)=2001200326+212+220+...+2�(�+1)=20012003
2.(12.3+13.4+14.5+...+1x(x+1))=200120032.(12.3+13.4+14.5+...+1�(�+1))=20012003
12−13+13−14+14−15+...+1x−1x+1=20012003:212−13+13−14+14−15+...+1�−1�+1=20012003:2
12−1x+1=2001400612−1�+1=20014006
=> 1x+1=12−20014006=120031�+1=12−20014006=12003
=> x + 1 = 2003
=> x = 2003 - 1
=> x = 2002
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