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mk gợi ý thui nhé :
cộng 96 phân số theo từng cặp:
a/b = (1/1+1/96)+(1/2+1/95)+(1/3+1/94)+...+(1/48+1/49)
...........................v.v
tự làm nhé
A=(1+1/96)(1/2+1/95).......................(1/48+1/49)
<=>A=97/96+97/190.........................97/2352
<=>A=97(1/96 x 1/190 x .................. x 1/2352)\(⋮97\)
=>A\(⋮97\)
k cho em mình nhé!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
a) x + 1/5 + x + 1/7 = x + 1/9
<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9
<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9
<=> 12/35x + 12/35 = 1/9x + 1/9
<=> 12/35x + 12/35 - 1/9x = 1/9
<=> 73/315x + 12/35 = 1/9
<=> 73/315x = 1/9 - 12/35
<=> 73/315x = -73/315
<=> x = 73/315 : -73/315 = -1
=> x = -1
b) làm tương tự
\(=\left(1+\frac{1}{96}\right)+\left(\frac{1}{2}+\frac{1}{95}\right)+...+\left(\frac{1}{47}+\frac{1}{48}\right)\)
\(=\frac{97}{1.96}+\frac{97}{2.95}+...+\frac{97}{47.48}\)
\(=97.\left(\frac{1}{1.96}+\frac{1}{2.95}+.....+\frac{1}{47.48}\right)\)
Vì 97 là số nguyên tố nên khi đưa về phân số tối giản
số đó sẽ có dạng \(\frac{97.A}{B}\left(B⋮97̸\right)\)
Vậy tử sẽ chia hết cho 97
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
\(\Rightarrow B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)}=\frac{1}{100}\)