tinh nhanh: B=1+1/3+1/5+...+1/97+1/99 / 1/1.99+1/3.97+1/5.95+...+1/49.51
Cho A = (1+1/3+1/5+...+1/97+1/99):(1/(1.99)+1/(3.97)+1/(5.95)+...+1/(49.51)). Rút gọn A
Ta thấy:
1/1 + 1/99 = (99+1)/(1.99)=100/(1.99)
1/3 + 1/97 = (97+3)/(3.97)=100/(3.97)
1/5 + 1/95 = (95+5)/(5.95)=100/(3.97)
…
1/97 + 1/3 = (3+97)/(97.3)=100/(97.3)
1/99 + 1/1 = (1+99)/(99.1)=100/(99.1)
=>
1/(1.99)=(1/1+1/99)/100
1/(3.97)=(1/3+1/97)/100
…
1/(99.1)=(1/99+1/1)/100
------------------------------ cộng 2 vế của các đẳng thức trên. Ta được đẳng thức:
1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 )
=[(1/1+1/99)+(1/3+1/99)+…+(1/99+1/1)]/1...
=2(1+1/3+1/5+1/7…+1/99]/100
=(1+1/3+1/5+1/7…+1/99]/50
Vậy:
A=(1+1/3+1/5+1/7+...+1/97+1/99) / [ 1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) ]
A=(1+1/3+1/5+1/7+...+1/97+1/99)/[(1+1/3...
A=50.
ko chắc nhé
bn ko lam theo cach lop 5 dc a . sao ma dan the
Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)
\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)
\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)
\(A=\frac{B}{6}=\frac{100}{2}=50\)
Vậy \(A=50\)
Tính nhanh \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)
Cho A=(1+1/3+1/5+...+1/97+1/99)/(1/1*99+1/3*97+1/5*95+...+1/49*51) rut gon ta duoc A=3
Không phải A=3 đâu bạn !!! Sau đây là cách giải của mik: Ta xét tử số : Đặt B=1+1/3+1/5+...+1/97+1/99 =>B=(1+1/99)+(1/3+1/97)+(1/5+1/95)+...+(1/49+1/51) =>B=100/1*99+100/3*97+100/5*95+...+100/49*51 =>B=100*(1/1*99+1/3*97+1/5*95+...+1/49*51) Ta có : A=B/(1/1*99+1/3*97+1/5*95+...+1/49*51) =>A=100*(1/1*99+1/3*97+1/5*95+...+1/49*51)/ (1/1*99+1/3*97+1/5*95+...+1/49*51) =>A=100 Vậy A=100 Mik chắc chắn 100% lun đó !!! Nếu các bạn thấy cách giải của mik hay thì nhớ *** cho mik nha (^.^) Thank you các bạn nhìu nhìu lắm ... >-<
\(\frac{1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+....+\frac{1}{49.51}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+......+\frac{1}{49.51}}\)
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{100\left(\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}\right)}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{\frac{1+99}{1.99}+\frac{3+97}{3.97}+\frac{5+95}{5.95}+...+\frac{49+51}{49.51}}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{\left(\frac{1}{1}+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)}\)
= \(\frac{100\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{51}+...+\frac{1}{99}}\)
= 100
\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)= ?
Cho A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}.\)Rút gọn A ta được A=.............
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)Rút gọn A ta được A =
Rút gọn \(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)
Giải
\(A=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\)
Cho A=\(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{49.51}}\) . Rút gọn A=