hay so sanh A=\(\frac{3^{2003}+5}{3^{2001+5}}\)va B=\(\frac{3^{2001}+1}{3^{1999}+1}\)
Những câu hỏi liên quan
Hay so sanh A=\(\frac{3^{2003+5}}{3^{2001}+5}\)va B=\(\frac{3^{2001+1}}{3^{1999+1}}\)
Khong quy đong mau so hay xo sanh cac phan so sau 2000/2001và 2002/2003 b 1998/1999và 1999/2000 c 2001/2000và 2002/2001 d 3/5 và 4/9 e 15/17và 123/127
Xem chi tiết
Tìm x :
a) \(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+ 3}{1998}+\frac{x+4}{1997}=-4\)
\(b.\frac{x+1}{1999}+\frac{x+2}{2000}+\frac{x+3}{2001}=\frac{x+4}{2002}+\frac{x+5}{2003}+\frac{x+6}{2004}\)
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
Đúng 0
Bình luận (0)
dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Đúng 0
Bình luận (0)
\(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+3}{1998}+\frac{x+4}{1997}=-4\)
\(\Leftrightarrow\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\) \(\left(\frac{x+4}{1997}+1\right)=0\)
\(\Leftrightarrow\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(\Leftrightarrow\left(x+2001\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
Mà : \(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\ne0\)
\(\Rightarrow x+2001=0\)
\(\Leftrightarrow x=-2001\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
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
Chứng minh rằng: frac{1}{5}+frac{1}{13}+frac{1}{25}+.....+frac{1}{1985} frac{9}{20}
so sánh A frac{3^{2003}+5}{3^{2001}+5}và B frac{3^{2001}+1}{3^{1009}+1}
so sánh A frac{1}{1.1981}+frac{1}{2.1982}+......+frac{1}{24.2004}+frac{1}{25.2005}
B frac{1}{1.61}+frac{1}{2.27}+.....+frac{1}{1979.2004}+frac{1}{1980.2005}
Các bn lm giúp mk nhé (dấu . là dấu x nhé )
Đọc tiếp
Chứng minh rằng: \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+.....+\frac{1}{1985}< \frac{9}{20}\)
so sánh A= \(\frac{3^{2003}+5}{3^{2001}+5}\)và B= \(\frac{3^{2001}+1}{3^{1009}+1}\)
so sánh A= \(\frac{1}{1.1981}+\frac{1}{2.1982}+......+\frac{1}{24.2004}+\frac{1}{25.2005}\)
B =\(\frac{1}{1.61}+\frac{1}{2.27}+.....+\frac{1}{1979.2004}+\frac{1}{1980.2005}\)
Các bn lm giúp mk nhé (dấu . là dấu x nhé )
Chứng minh rằng: frac{1}{5}+frac{1}{13}+frac{1}{25}+.....+frac{1}{1985} frac{9}{20}so sánh A frac{3^{2003}+5}{3^{2001}+5}và B frac{3^{2001}+1}{3^{1009}+1}so sánh A frac{1}{1.1981}+frac{1}{2.1982}+......+frac{1}{24.2004}+frac{1}{25.2005} B frac{1}{1.61}+frac{1}{2.27}+.....+frac{1}{1979.2004}+frac{1}{1980.2005}Các bn lm giúp mk nhé (dấu . là dấu x nhé )
Đọc tiếp
Chứng minh rằng: \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+.....+\frac{1}{1985}< \frac{9}{20}\)
so sánh A= \(\frac{3^{2003}+5}{3^{2001}+5}\)và B= \(\frac{3^{2001}+1}{3^{1009}+1}\)
so sánh A= \(\frac{1}{1.1981}+\frac{1}{2.1982}+......+\frac{1}{24.2004}+\frac{1}{25.2005}\)
B =\(\frac{1}{1.61}+\frac{1}{2.27}+.....+\frac{1}{1979.2004}+\frac{1}{1980.2005}\)
Các bn lm giúp mk nhé (dấu . là dấu x nhé )
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,\(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right):x=\frac{-2001}{2002}\).
Giải phương trình
\(\frac{x+1}{2003}+\frac{x+3}{2001}+\frac{x+5}{1999}-\frac{x+7}{1997}-\frac{x+9}{1995}-\frac{x+11}{1993}=0\)
Tính B, biết:
B=\(\frac{1}{3+5}+\frac{1}{5+7}+...+\frac{1}{1999+2001}\)
Đề sai rồi . Xem lại đề đi
Đúng 0
Bình luận (0)