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Cách làm :
Áp dụng công thức : \(\dfrac{n}{a\left(a+n\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\)
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=1-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2F=\dfrac{2}{1.3}+\dfrac{2}{3.5}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+........+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2F=1-\dfrac{1}{101}\)
\(\Leftrightarrow2F=\dfrac{100}{101}\)
\(\Leftrightarrow F=\dfrac{50}{101}\)
Giải:
\(C=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{999.1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{999}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{1}{1}-\dfrac{1}{1000}\)
\(\Leftrightarrow C=\dfrac{999}{1000}\)
Sửa đề:
\(F=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{999.1001}\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{999}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{1001}\right)\)
\(\Leftrightarrow F=\dfrac{1}{2}.\dfrac{1000}{1001}\)
\(\Leftrightarrow F=\dfrac{500}{1001}\)
Chúc bạn học tốt!
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2020.2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=1-\frac{1}{2021}=\frac{2020}{2021}\)
b) \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{21.23}=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{21.23}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{21}-\frac{1}{23}\right)=\frac{1}{2}\left(1-\frac{1}{23}\right)=\frac{1}{2}.\frac{22}{23}=\frac{11}{23}\)
c) \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}=\frac{1}{99}-\left(\frac{1}{98.99}+\frac{1}{97.98}+...+\frac{1}{1.2}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}+\frac{1}{97}-\frac{1}{98}+...+1-\frac{1}{2}\right)=\frac{1}{99}-\left(-\frac{1}{99}+1\right)=\frac{1}{99}-\frac{98}{99}\)
\(=-\frac{97}{99}\)
d) bạn xem lại đề
a)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)
\(=\frac{1}{1}-\frac{1}{2021}\)
\(=\frac{2020}{2021}\)
b)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{21\cdot23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{23}\right)\)
\(=\frac{1}{2}\cdot\frac{22}{23}\)
\(=\frac{11}{23}\)
c)
\(=\frac{1}{99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{99}-\frac{98}{99}\)
\(=\frac{-97}{99}\)
d)
đề sai hay sao á mong bạn xem ljai ạ
\(\text{Đ}\text{ặt}:A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{99.101}\)
\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(2A=1-\frac{1}{101}\)
\(A=\frac{100}{101}:2=\frac{50}{101}\)
\(\Rightarrow\frac{1}{3}x.x=\frac{50}{101}\)
\(x.\left(\frac{1}{3}.1\right)=\frac{50}{101}\)
\(x.\frac{1}{3}=\frac{50}{101}\)
$x=\frac{50}{101}:\frac{1}{3}=\frac{150}{101}$
\(.\frac{1}{3}x.x=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\frac{1}{3}xx=\frac{1}{2}.\left(\frac{100}{101}\right)\)
\(\frac{1}{3}xx=\frac{50}{101}\)
\(x.x=\frac{150}{101}\)
còn lại tự tính
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(\Rightarrow2A=1-\frac{1}{101}\)
\(\Rightarrow2A=\frac{100}{101}\)
\(\Rightarrow A=\frac{100}{101}:2\)
\(\Rightarrow A=\frac{50}{101}.\)
Chúc bạn học tốt!
\(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\\ A=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right)\\ A=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\\ A=\frac{1}{2}\left(1-\frac{1}{101}\right)\\ A=\frac{1}{2}\cdot\frac{100}{101}\\ A=\frac{50}{101}\)
\(B=1-2+3-4+...+49-50\\ B=\left(1-2\right)+\left(3-4\right)+...+\left(49-50\right)\\ B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\text{ (có 25 số -1)}\\ B=\left(-1\right)\cdot25=-25\)
Đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
\(2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...+\frac{1}{99}-\frac{1}{101}\)
\(2A=\frac{100}{101}\)
\(A=\frac{50}{101}\)
b) \(\frac{2^{10}+3^{31}+2^{40}+3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}=\frac{2^{10}+2^{40}}{2^{11}+2^{41}}\)
\(\frac{2^{10}+2^{40}}{2^{11}+2^{41}}=\frac{1}{2}\)
=1/2x(1/1.3+1/3.5+...+1/99.101)
=1/2.(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)
=1/2.(1-1/101)
=1/2.100/101
=50/101
chúc bạn học tốt
\(S=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
A = 1/1.2 + 1/2.3 + 1/3.4 + .... + 1/99.100
A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.....+ 1/99- 1/100
A= 1 - 1/100
A= 99/100
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