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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=1.\left(1-\frac{1}{100}\right)\)
\(=1.\frac{99}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Hình như là fan BTS ms đúng mà~
1/1 . 2 + 1/ 3 . 4 + 1/5 . 6 + ...+ 1/99 . 100
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/99 - 1/100
= ( 1 + 1/3 + 1/5 + ...+ 1/99 ) - ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - 2 . ( 1/2 + 1/4 + ...+ 1/100 )
= ( 1 + 1/2 + 1/3 + ...+ 1/99 + 1/100 ) - ( 1 + 1/2 + ...+ 1/50 )
= 1/51 + 1/52 + ...+ 1/100
Tham khảo nha !!!
\(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{99\cdot100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}-1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\) (đpcm)
đặt A = 1/1*2 + 1/3*4 + 1/5*6 + ... + 1/99*100
= 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/99 - 1/100
= (1 + 1/3 + 1/5 + ... + 1/99) - (1/2 + 1/4 + 1/6 + ... + 1/100)
= 1 + 1/2 + 1/3 + ... + 1/100 - 2(1/2 + 1/4 + 1/6 + .... + 1/100)
= 1 + 1/2 + 1/3 + ... + 1/100 - 1 - 1/2 - 13 - ... - 1/50
= 1/51 + 1/52 + 1/53 + ... + 1/100
thay vào ra E = 1
Biến đổi mẫu ta được:
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(\Rightarrow E=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}=1\)
đặt A =1/1.2+1/3.4+1/5.6+...+1/99.100
A=1/1-1/2+1/3-1/4+...+1/99-1/100
A=1+1/2+1/3+...+1/100-2(1/2+1/4+...+1/100)
A=1+1/2+1/3+...+1/100-1-1/2-1/3-...-1/50
A=1/51+1/52+...+1/100
=>D=1
D là dãy số đã cho nha bn
nếu đúng thì k nha
Tính $E=\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..+\frac{1}{99.100}}$E=151 +152 +153 +....+1100 11.2 +13.4 +15.6 +..+199.100
Toán lớp 6
Rút gọn mẫu ta được:
\(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}\)
Vì tử và mẫu bằng nhau nên biểu thức bằng 1
Bạn muốn biết cách rút gọn mẫu thì gửi tin nhắn cho mình
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{2-1}{1.2}+\frac{4-3}{3.4}+\frac{6-5}{5.6}+...+\frac{100-99}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{99}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=> \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=> A = 1
a=\(1-\frac{1}{2}+\frac{1}{3}-...-\frac{1}{100}=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)
=>b/a=2011
hình như đề : CMR : \(\frac{b}{a}\)là 1 số nguyên
Ta có :
\(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(a=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(a=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(a=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)
\(a=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(a=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(b=\frac{2011}{51}+\frac{2011}{52}+\frac{2011}{53}+...+\frac{2011}{100}\)
\(b=2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)\)
\(\Rightarrow\frac{b}{a}=\frac{2011.\left(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\right)}{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}}=2011\)là 1 số nguyên ( đpcm )
uh.cậu là fan của bts hả.mình cũng thế,nhưng mình thích red velvet hơn
Biến đổi VT ta có :
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{50}\right)\)
\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}=VP\)
\(\Rightarrowđpcm\)