Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
Chứng minh 1 bất đẳng thức cơ bản sau:\(\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2n\left(n+1\right)}\)
Thật vậy: \(\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{n^2+n^2+2n+1}=\dfrac{1}{2n^2+2n+1}< \dfrac{1}{2n^2+2n}=\dfrac{1}{2n\left(n+1\right)}\)
Thay vào bài toán \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+\dfrac{1}{3^2+\left(3+1\right)^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{2n\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}-\dfrac{1}{2\left(n+1\right)}< \dfrac{1}{2}\left(đpcm\right)\)
\(=\dfrac{\left(13+\dfrac{1}{4}-2-\dfrac{5}{27}-10-\dfrac{5}{6}\right)\cdot230.04+46.75}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\dfrac{37}{3}-14-\dfrac{2}{7}}\)
\(=\dfrac{\dfrac{25}{108}\cdot\dfrac{5751}{25}+46.75}{\dfrac{100}{21}\cdot\dfrac{3}{37}-\dfrac{100}{7}}\)
\(=\dfrac{100}{\dfrac{-3600}{259}}=-\dfrac{259}{36}\)
\(a,=\dfrac{13}{50}\cdot\dfrac{50}{13}\cdot\left(-\dfrac{31}{2}\right)\cdot\dfrac{169}{2}=-\dfrac{5239}{2}\\ b,=\dfrac{-\dfrac{49}{100}\cdot\left(-125\right)}{-\dfrac{343}{27}\cdot\dfrac{81}{16}\cdot\left(-1\right)}=\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{245}{4}\cdot\dfrac{16}{1029}=\dfrac{20}{21}\)
a) \(\dfrac{13}{50}.\left(-15.5\right):\dfrac{13}{50}.84\dfrac{1}{2}=\dfrac{13}{50}.-75:\dfrac{13}{50}.\dfrac{169}{2}=-\dfrac{75.169}{2}=-\dfrac{12675}{2}\)
b) \(\dfrac{\left(-0,7\right)^2.\left(-5\right)^3}{\left(-2\dfrac{1}{3}\right)^3.\left(1\dfrac{1}{2}\right)^4.\left(-1\right)^5}=\dfrac{0,49.\left(-125\right)}{-\dfrac{343}{27}.\dfrac{81}{16}.\left(-1\right)}=-\dfrac{\dfrac{245}{4}}{\dfrac{1029}{16}}=\dfrac{20}{21}\)
Ta dễ dàng chứng minh được: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\)
Thật vậy:
\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1>2n^2+2n=2n\left(n+1\right)\)Trở lại bài toán
\(A=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(A=\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+....+\dfrac{1}{n^2+\left(n+1\right)^2}\)
\(A< \dfrac{1}{2.1.\left(1+1\right)}+\dfrac{1}{2.2.\left(2+1\right)}+\dfrac{1}{2.3.\left(3+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\)
\(A< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{2n\left(n+1\right)}\)
\(A< \dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}\right)\)
\(A< \dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(A< \dfrac{1}{2}\left(1-\dfrac{1}{n+1}\right)\)
\(A< \dfrac{1}{2}-\dfrac{1}{2n+2}< \dfrac{1}{2}\left(đpcm\right)\)