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\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+.....+\frac{n^2-1}{n^2}\)
\(=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+....+\frac{n^2-1}{n^2}\)
\(=\left(1+1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{n^2}\right)\)
\(=n-1-\left(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{4^2}\right)\)
Mà \(0< \frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{4^2}< 1\) ( không biết chứng minh thì ib )
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{4^2}\) không là số nguyên => đpcm
b,\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n.\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(\Rightarrow D=1-\frac{1}{n+2}=\frac{n}{n+2}< \frac{n+2}{n+2}=1\left(1\right)\)
\(\Rightarrow D=\frac{n}{n+2}>0\left(2\right)\)
Từ (1);(2)\(\Rightarrow0< D< 1\)
\(\Rightarrowđpcm\)
a,\(C>0\)
\(C=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}< 9;\frac{1}{11}< 1\)
\(\Rightarrow0< A< 1\)
\(\Rightarrow A\notinℤ\)
c,\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Ta quy đồng 3 số đầu
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>\frac{6.2}{12}=1\)
\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< \frac{6.2}{6}=2\)
\(1< E< 2\)
\(E\notinℤ\)
1b) Ta có: \(\frac{x}{3}\) = \(\frac{y}{4}\) => \(\frac{x}{15}\) = \(\frac{y}{20}\)
\(\frac{y}{5}\) = \(\frac{z}{6}\) => \(\frac{y}{20}\) = \(\frac{z}{24}\)
=> \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\)
Đặt \(\frac{x}{15}\) = \(\frac{y}{20}\) = \(\frac{z}{24}\) = k
=> x = 15k; y = 20k và z = 24k
Thay vào A ta có:
A = \(\frac{2.15k+3.20k+4.24k}{3.15k+4.20k+5.24k}\)
=> A = \(\frac{30k+60k+96k}{45k+80k+120k}\)
=> A = \(\frac{\left(30+60+96\right)k}{\left(45+80+120\right)k}\)
=> A = \(\frac{186k}{245k}\)
=> A = \(\frac{186}{245}\)
Vậy A = \(\frac{186}{245}\).
\(N=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
=>\(N=\frac{13860}{41580}+\frac{10385}{41580}+\frac{8316}{41580}+\frac{11880}{41580}+\frac{9240}{41580}+\frac{7560}{41580}\)
=>\(N=\frac{61251}{41580}\)
=>N ko phải là số nguyên (đpcm)
HỌC TÔT :)
\(=\frac{2\cdot4}{3^2}\cdot\frac{3.5}{4^2}\cdot\frac{4\cdot6}{5^2}\cdot......\cdot\frac{49\cdot51}{50^2}\)
=\(\frac{\left[2\cdot3\cdot4\cdot......\cdot49\right]\cdot\left[4\cdot5\cdot6\cdot.....\cdot51\right]}{\left[3\cdot4\cdot5\cdot....\cdot50\right]\cdot\left[3\cdot4\cdot5\cdot....\cdot50\right]}\)
=\(\frac{2\cdot51}{50\cdot3}\)
=\(\frac{17}{25}\)
Vì \(\frac{17}{25}\) ko phải là số nguyên nên B ko phải là số nguyên [ĐPCM]
Đặt A =\(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)\)
Đặt B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
>\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{2}-\frac{1}{101}=\frac{99}{202}\)
Khi đó A = \(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)< 99-\frac{99}{202}\approx98,5\)
=> A < 98,5 (1)
Lại có B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
Khi đó A =\(99-\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\right)>99-\frac{99}{100}=98,01\)
=> A > 98,01 (2)
Từ (1)(2) => 98,01 < A < 98,5
=> A không là số nguyên