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\(=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)
=>A=\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
Và đến đây là hết biik giải nữa
Ta có:
1/49 + 1 = 50/49
2/48 + 1 = 50/48
3/47 + 1 = 50/47
.
.
.
47/3 + 1 = 50/3
48/2 + 1 = 50/2
0 + 1 = 50/50
Cộng vế theo vế dãy đẳng thức trên ta được:
1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50/2 + 50/3 + 50/4 +........+ 50/49 + 50/50
⇒ 1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50 x (1/2 + 1/3 + 1/4 +........+ 1/49 + 1/50)
⇒ B = 50A
⇒ A/B = 1/50
Ta có:
1/49 + 1 = 50/49
2/48 + 1 = 50/48
3/47 + 1 = 50/47
.
.
.
47/3 + 1 = 50/3
48/2 + 1 = 50/2
0 + 1 = 50/50
Cộng vế theo vế dãy đẳng thức trên ta được:
1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50/2 + 50/3 + 50/4 +........+ 50/49 + 50/50
⇒ 1/49 + 2/48 +........+ 47/3 + 48/2 + 49 = 50 x (1/2 + 1/3 + 1/4 +........+ 1/49 + 1/50)
⇒ B = 50A
⇒ A/B = 1/50
đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}\)
\(=\frac{100}{1}-1+\frac{100}{2}-1+...+\frac{100}{99}-1=\left(\frac{100}{1}+\frac{100}{2}+...+\frac{100}{99}\right)-\left(1+1+...+1\right)\)
\(100+\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}\right)-99=1+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}}=\frac{B}{100B}=\frac{1}{100}\)
đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}=\frac{100}{1}-1+\frac{100}{2}-1+...+\frac{100}{99}-1\)
\(=\left(\frac{100}{1}+\frac{100}{2}+...+\frac{100}{99}\right)-\left(1+1+...+1\right)=100+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-99\)
\(=1+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100B\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}}=\frac{B}{100B}=\frac{1}{100}\)
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Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\)
\(=\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+\left(\frac{1}{3}+\frac{1}{96}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\)
\(=\frac{99}{1.98}+\frac{99}{2.97}+\frac{99}{3.96}+...+\frac{99}{49.50}\)
\(=99\left(\frac{1}{1.98}+\frac{1}{2.97}+\frac{1}{3.96}+...+\frac{1}{49.50}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{98}\right).2.3.4....98\)
\(=99\left(\frac{1}{1.98}+\frac{1}{2.97}+\frac{1}{3.96}+...+\frac{1}{49.50}\right).2.3.4....98\)chia hết cho 99 (đpcm)
a) Đặt B = \(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)
\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)
\(=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}\)
\(=100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\right)\)
Đặt C = \(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{99.1}\)
\(=\left(\frac{1}{1.99}+\frac{1}{99.1}\right)+\left(\frac{1}{3.97}+\frac{1}{97.3}\right)+...+\left(\frac{1}{49.51}+\frac{1}{51.49}\right)\)
\(=2\cdot\frac{1}{1.99}+2\cdot\frac{1}{3.97}+...+2\cdot\frac{1}{49.51}\)
\(=2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)\)
Thay B và C vào A
\(\Rightarrow A=\frac{100\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}{2\left(\frac{1}{1.99}+\frac{1}{3.97}+...+\frac{1}{49.51}\right)}=\frac{100}{2}=50\)
b) Đặt E = \(\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}\)
\(=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\)
\(=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Thay E vào B
\(\Rightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
100A = \(\frac{99}{1}+1+\frac{98}{2}+1+...+\frac{1}{99}+1-99\)
100A=\(\frac{100}{1}+\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}-99\)
100A =\(\left(\frac{100}{2}+\frac{100}{3}+..+\frac{100}{99}+100-99\right)\)
100A=\(\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\right)\)
100A=\(\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+\frac{100}{100}\right)\)
100A=100.\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
A=\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)