1/3+13/15+33/35+31/63+.....................+9601/9603+9997/9999
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\(\frac{1}{3}+\frac{13}{15}+...+\frac{9997}{9999}\)
\(=1-\frac{2}{3}+1-\frac{2}{15}+...+1-\frac{2}{9999}\)
\(=\left(1+1+...+1\right)-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=50-\left(1-\frac{1}{101}\right)\)
Sau bạn tính tiếp là OK rồi
\(\dfrac{1}{3}+\dfrac{13}{15}+\dfrac{33}{35}+...+\dfrac{9997}{9999}\)
\(=1-\dfrac{2}{3}+1-\dfrac{2}{15}+1-\dfrac{2}{35}+...+1-\dfrac{2}{9999}\)
\(=\left(1+1+1+...+1\right)-\dfrac{2}{3}+\dfrac{2}{15}+...+\dfrac{2}{9999}\)
\(=50-1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(=50-\left(1-\dfrac{1}{101}\right)=50-\dfrac{100}{101}\)
\(=\dfrac{4950}{101}\)
Đặt \(A=\)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{143}\)
\(=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{11.13}\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\)
\(2A=\frac{1}{3}-\frac{1}{13}=\frac{10}{39}\)
\(A=\frac{5}{39}\)
Câu còn lại cx dựa như vậy nhé bn !
Chúc bn hc tốt <3
3/15+3/35+3/63+...+3/9603
=3/3.5+3/5.7+3/7.9+...+3/97.99
=3/2.(2/3.5+2/5.7+2/7.9+...+2/97.99)
=3/2.(1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99)=3/2.(1/3-1/99)
=3/2.32/99
=16/33
3/15+3/35+3/63+.....+3/9603
=3/3.5+3/5.7+3/7.9+.....+3/97.99
=5-3/3.5+7-5/5.7+9-7/7.9+.....+99-97/97.99
=5/3.5-3/3.5+7/5.7-5/5.7+9/7.9-7/7.9+.....+99/97.99-97/97.99
=1/3-1/5+1/5-1/7+1/7-1/9+.....+1/97-1/99
=1/3-1/99
=32/99
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