Tính nhanh.
a) \(\frac{2003x14+1988+2001+2002}{2002+2002x503+504x2002}\)
b) \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{28}\)
Những câu hỏi liên quan
Tính nhanh: \(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)
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Lưu ý: dấu ' . ' là dấu ' x '.
\(\frac{2003.14+1988+2001.2002}{2002+2002.503+504.2002}\)
\(=\frac{2002.2001+2002.14+14+1988}{2002.\left(503+504+1\right)}\)
\(=\frac{2002.2001+2002.14+2002}{2002.1008}\)
\(=\frac{2002.\left(2001+14+1\right)}{2002.1008}\)
\(=\frac{2002.2016}{2002.1008}\)
\(=\frac{2002.2^4.3^2.7.2}{2002.2^4.3^2.7}\)
\(=2\)
Xem thêm câu trả lời
Tính nhanh
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)
Tính nhanh
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)= \(\left(\frac{2003x14+1988}{2002+2002x503}\right)+\frac{2001x2002}{504x2002}\) = \(\frac{5}{168}+\frac{667}{168}\)= \(\frac{672}{168}=4\)
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Bài phân số mới ra lò đuê:
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}\)
Nhanh, đúng, rõ ràng mk tick
\(\frac{2003x14+1988+2001x2002}{2002+2002x503+504x2002}=\frac{2002x14+14+1988+2001x2002}{2002x\left(1+503+504\right)}\)
\(=\frac{2002x14+2002+2001x2002}{2002x1008}=\frac{2002x\left(14+1+2001\right)}{2002x1008}\)
\(=\frac{2002x2016}{2002x1008}=2\)
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Tính nhanh \(\frac{2003x14+1988+2001x2002}{2002+2004x503+504x2002}\)
Tính nhanh
\(\frac{2003x14+1988+2001x2002}{2002+2004x503+504x2002}\)
Làm cả cách làm
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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chứng minh : \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....-\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+.....+\frac{1}{2002}\)
\(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)
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CMR:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{\text{4}}+...+\frac{1}{2001}-\frac{1}{2002}=\frac{1}{1002}+...+\frac{1}{2002}\)
Ta có \(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+...\frac{1}{2002}=VP\)
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