Giải thích các bước giải:
Câu 1:
Ta có:
$B=\dfrac{99}{1}+\dfrac{98}{2}+...+\dfrac{1}{99}$
$\to B=99+\dfrac{98}{2}+...+\dfrac{1}{99}$
$\to B=(1+\dfrac{98}{2})+(1+\dfrac{97}{3})+...+(1+\dfrac1{99})+1$
$\to B=\dfrac{2+98}{2}+\dfrac{3+97}{3}+...+\dfrac{99+1}{99}+1$
$\to B=\dfrac{100}{2}+\dfrac{100}{3}+...+\dfrac{100}{99}+1$
$\to B=100(\dfrac12+\dfrac13+...+\dfrac1{99}+\dfrac1{100})$
$\to B=100A$
$\to \dfrac{A}{B}=\dfrac{1}{100}$
Bài 2:
Ta có:
$A=\dfrac{1+(1+2)+(1+2+3)+...+(1+2+...+98)}{1.98+2.97+...+98.1}$
$\to A=\dfrac{1+\dfrac{2(2+1)}{2}+\dfrac{3(3+1)}{2}+...+\dfrac{98(98+1)}{2}}{1.98+2.97+...+98.1}$
$\to A=\dfrac{1+\dfrac{2.3}{2}+\dfrac{3.4}{2}+...+\dfrac{98.99}{2}}{1.98+2.97+...+98.1}$
$\to A=\dfrac{\dfrac12(1.2+2.3+3.4+...+98.99)}{1.98+2.97+...+98.1}$
$\to A=\dfrac12\cdot \dfrac{1.2+2.3+3.4+...+98.99}{1.98+2.97+...+98.1}$
Mà $1.2+2.3+3.4+...+98.99$
$=1.(100-98)+2.(100-97)+3.(100-96)+...+98.(100-1)$
$=1.100-1.98+2.100-2.97+3.100-3.96+...+98.100-98.1$
$=(1.100+2.100+3.100+...+98.100)-(1.98+2.97+3.96+...+98.1)$
$=100(1+2+3+..+98)-(1.98+2.97+3.96+...+98.1)$
$=100\cdot \dfrac{98(98+1)}{2}-(1.98+2.97+3.96+...+98.1)$
$=100\cdot \dfrac{98.99}{2}-(1.98+2.97+3.96+...+98.1)$
$=\dfrac{98.99.100}{2}-(1.98+2.97+3.96+...+98.1)$
Lại có:
$B=1.2+2.3+3.4+...+98.99$
$\to 3B=1.2.3+2.3.3+3.4.3+...+98.99.3$
$\to 3B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+98.99.(100-97)$
$\to 3B=-0.1.2+1.2.3-1.2.3+2.3.4-2.3.4+3.4.5+...-97.98.99+98.99.100$
$\to 3B=98.99.100$
$\to B=\dfrac{98.99.100}{3}$
$\to \dfrac{98.99.100}{3}=\dfrac{98.99.100}{2}-(1.98+2.97+3.96+...+98.1)$
$\to 1.98+2.97+3.96+...+98.1=\dfrac{98.99.100}{6}$
$\to A=\dfrac12\cdot \dfrac{\dfrac{98.99.100}{3}}{\dfrac{98.99.100}{6}}$
$\to A=1$
