Tính B, biết:
B=\(\frac{1}{3+5}+\frac{1}{5+7}+...+\frac{1}{1999+2001}\)
NM
Những câu hỏi liên quan
Tìm x, biết :a, left(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{98cdot99cdot100}right)x-3;b, left(frac{frac{2000}{1}+frac{1999}{2}+...+frac{1}{2000}}{frac{1}{2}+frac{1}{3}+...+frac{1}{2001}}right)xfrac{-1}{5}.c,left(frac{frac{2000}{1}+frac{1999}{2}+...+frac{1}{2000}+2000}{1+frac{1}{2}+frac{1}{3}+...+frac{1}{2001}}right):xfrac{-2001}{2002}.
Đọc tiếp
Tìm x, biết :
a, \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\right)x=-3\);
b, \(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right)x=\frac{-1}{5}\).
c,\(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right):x=\frac{-2001}{2002}\).
Hay so sanh A=\(\frac{3^{2003+5}}{3^{2001}+5}\)va B=\(\frac{3^{2001+1}}{3^{1999+1}}\)
hay so sanh A=\(\frac{3^{2003}+5}{3^{2001+5}}\)va B=\(\frac{3^{2001}+1}{3^{1999}+1}\)
Tính
A bằng 1+2-3-4+5+6-7-8+.......-1999-2000+2001+2002-2003
B bằng \(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)......\left(\frac{1}{121}-1\right)\)
Gộp nhóm 4 => A = -4 * 500+2001+2002-2003=0
B =
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a) tạm bỏ số 1 ra => có 2012 số hạng=> có 1006 cặp =(-1)
=> A=1+-(-1).1006=-1005
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1/ Tính : \(\frac{-8}{5}+\frac{207207}{201201}\)
2/ Tính:
\(M=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}}{\frac{2001}{1}+\frac{2002}{2}+\frac{1999}{3}+...+\frac{1}{2001}}\)
1)\(\frac{-8}{5}+\frac{207207}{201201}\)
=\(\frac{-8}{5}+\frac{207}{201}\)
=\(\frac{-8}{5}+\frac{69}{67}\)
=\(\frac{-191}{335}\)
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Giải phương trình
\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)
\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}=\frac{x+7}{1997}+\frac{x+9}{1995}+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}-\frac{x+2004}{1997}-\frac{x+2004}{1995}-\frac{x+2004}{1993}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\) ( do \(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}+\frac{1}{1997}+\frac{1}{1995}+\frac{1}{1993}\ne0\))
\(\Leftrightarrow x=-2004\)
\(\frac{x+1}{2003}\)\(+\)\(\frac{x+3}{2001}\)\(+\)\(\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+\frac{x+9}{1995}\)\(+\frac{x+11}{1993}\)
\(\Leftrightarrow\)\(\frac{x+1}{2003}\)\(+1+\)\(\frac{x+3}{2001}\)\(+1+\frac{x+5}{1999}\)= \(\frac{x+7}{1997}\)\(+1+\frac{x+9}{1995}\)\(+1+\frac{x+11}{1993}\)
\(\Leftrightarrow\frac{x+2004}{2003}\)\(+\frac{x+2004}{2001}\)\(+\frac{x+2004}{1999}\)\(-\frac{x+2004}{1997}\)\(-\frac{x+2004}{1995}\)\(-\frac{x+2004}{1993}\)\(=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
\(\Leftrightarrow x+2004=0\)(vì tích kia có kết quả khác 0)
\(\Leftrightarrow x=-2004\)
Vậy PT có tập nghiệm S = {-2004}
\(\frac{x+1}{2003}+1+\frac{x+3}{2001}+1+\frac{x+5}{1999}+1=\frac{x+7}{1997}+1+\frac{x+9}{1995}+1+\frac{x+11}{1993}+1\)
<=>\(\frac{x+2004}{2003}+\frac{x+2004}{2001}+\frac{x+2004}{1999}=\frac{x+2004}{1997}+\frac{x+2004}{1995}+\frac{x+2004}{1993}\)
<=>\(\left(x+2004\right)\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)=0\)
Mà \(\left(\frac{1}{2003}+\frac{1}{2001}+\frac{1}{1999}-\frac{1}{1997}-\frac{1}{1995}-\frac{1}{1993}\right)< 0\)
=> X+2004=0
=>X=-2004
Giải phương trình
\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}-\frac{x+7}{1997}-\frac{x+9}{1995}-\frac{x+11}{1993}=0\)
Tìm x :
a) \(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+ 3}{1998}+\frac{x+4}{1997}=-4\)
\(b.\frac{x+1}{1999}+\frac{x+2}{2000}+\frac{x+3}{2001}=\frac{x+4}{2002}+\frac{x+5}{2003}+\frac{x+6}{2004}\)
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
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dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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\(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+3}{1998}+\frac{x+4}{1997}=-4\)
\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)
\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)
\(\Rightarrow x+2001=0\)
\(\Leftrightarrow x=-2001\)
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\(A=\left\{\frac{1999}{2011}-\frac{2011}{1999}\right\}-\left\{\frac{-12}{1999}-\frac{12}{2011}\right\}\)
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