1/3+13/15+33/35+61/62+...+9601/9603+9997/9999
Những câu hỏi liên quan
tính nhanh 1/3 + 13/15 + 33/35 + 61/65 +......+ 9601/9603+ 9997/9999
tính
1/3+13/15+33/35+61/63+...+9601/9603+9997/9999
tl nhanh jup mik, đag cần gấp
\(\frac{1}{3}+\frac{13}{15}+...+\frac{9997}{9999}\)
\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)
\(=\left(1+1+...+1\right)-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=50-\left(1-\frac{1}{101}\right)\)
Sau bạn tính tiếp là OK rồi
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Tìm M : M= 1/3+13/15+33/35+...+9601/9603+9997/9999
1/3+13/15+33/35+31/63+.....................+9601/9603+9997/9999
cứu với
1/3+13/15+33/35+31/63+.....................+9601/9603+9997/9999
\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)
\(=\left(1+1+1+1+...+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)
\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{4950}{101}\)
HTDT
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\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)
\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=50-\left(1-\dfrac{1}{101}\right)=50-\dfrac{100}{101}\)
\(=\dfrac{4950}{101}\)
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giải nhanh giúp mình câu dưới đây nhé thứ 3 ngày 22/8 mình phải đi học rồi
\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+.....+\frac{9997}{9999}\)
\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+...+\frac{9997}{9999}=1-\frac{2}{3}+1-\frac{2}{15}+1-\frac{2}{35}+...+1-\frac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)
\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{4950}{101}\)
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thank you bạn nhé mình sẽ k cho bạn
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nhưng mà sao bạn biết là có 50 số 1
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\(C=\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+....+\frac{9997}{9999}\)
Tìm C giúp mình nha! Các bạn giải theo cách cấp 1 nhé!
\(1005+\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9603}+\frac{1}{9999}\)
1005\(\frac{49}{303}\)
M=1+1/5+3/35+...+3/9603+3/9999
\(M=1+\frac{1}{5}+\frac{3}{35}+...+\frac{3}{9603}+\frac{3}{9999}\)
\(=\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}+\frac{3}{99\times101}\)
\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}+\frac{2}{99\times101}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{101}\right)=\frac{150}{101}\)