Lời giải:
Bài 1:
14)
$A=\dfrac{2^{12}. 3^5 - 4^6 . 9^2 }{(2^2 . 3)^6 + 8^4 . 3^5}- \dfrac{5^{10} . 7^3 -25^5 . 49^2}{(125.7)^3 + 5^9 . 14^3}$
$A=\dfrac{2^{12} . 3^5 -(2^2)^6 . (3^2)^2}{(2^2)^6 . 3^6 + (2^3)^4 . 3^5}- \dfrac{5^{10} . 7^3 - (5^2)^5 . 7^4}{(5^3)^3 . 7^3 + 5^9 . 14^3}$
$A=\dfrac{2^{12} . 3^5 -2^{12} . 3^9}{2^{12} . 3^6 + 2^{12} . 3^5}- \dfrac{5^{10} . 7^3 - 5^{10} . 7^4}{ 5^9 . 7^3 + 5^9 . 14^3}$
$A=\dfrac{2^{12} . (3-1). 3^4}{2^{12} .( 3^2 + 3) .3^4}- \dfrac{5^{9} . (5-5.7) . 7^3}{ 5^9 . ( 1+2^3).7^3}$
`A=1/6 - (-3/10)`
`A=7/15`
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15)
$A=\dfrac{3}{1^2 . 2^2}+\dfrac{5}{2^2 . 3^2}+\dfrac{7}{3^2 . 4^2}+...+\dfrac{19}{9^2 . 10^2}$
$A=\dfrac{2^2-1}{1^2 . 2^2}+\dfrac{3^2-2^2}{2^2 . 3^2}+\dfrac{4^2-3^2}{3^2 . 4^2}+...+\dfrac{10^2 - 9^2}{9^2 . 10^2}$
$A=\dfrac{1}{1} - \dfrac{1}{2^2} + \dfrac{1}{2^2} - \dfrac{1}{3^3} + \dfrac{1}{3^3} + ... + \dfrac{1}{9^2} - \dfrac{1}{10^2} $
$A=\dfrac{1}{1} + [- \dfrac{1}{2^2} + \dfrac{1}{2^2}]+[ - \dfrac{1}{3^3} + \dfrac{1}{3^3}] + ... + [-\dfrac{1}{9^2}+ \dfrac{1}{9^2}]- \dfrac{1}{10^2} $
$A=\dfrac{1}{1}-\dfrac{1}{100}$
`A-99/100`
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16)
$A=1+5+5^2 + 5^3 +...+ 5^{49} + 5^{50}$
$5A=5+5^2 + 5^3 + ...+ 5^{50} + 5^{51}$
$5A-A=(5+5^2 + 5^3 + ...+ 5^{50} + 5^{51})-(1+5+5^2 + 5^3 +...+ 5^{49} + 5^{50})$
$4A=5^{51}-1$
$A=\dfrac{5^{51}-1}{4}$
