\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)<\(\dfrac{1}{4}\)

Đặt A=\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\)

2.A=2.(\(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{17.18.19}\))

2. A=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+\(\dfrac{2}{3.4.5}\)+...+\(\dfrac{2}{17.18.19}\)

2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+ ...+\(\dfrac{1}{17.18}\)-\(\dfrac{1}{18.19}\)

2.A=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{18.19}\)=\(\dfrac{85}{171}\)

A=\(\dfrac{85}{171}\):2=\(\dfrac{85}{342}\)

Ta cũng có: \(\dfrac{1}{4}\) = \(\dfrac{171}{684}\); \(\dfrac{85}{342}\) = \(\dfrac{170}{684}\)

Vì 170 < 171 ( \(\dfrac{170}{684}\) < \(\dfrac{171}{684}\) )

Vậy \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{17.18.19}\) < \(\dfrac{1}{4}\)

8.x = 7,8 . x + 25

=>8.x-7,8.x = 25

x.(8-7,8) = 25

x . 0,2 =25

x = 25:0.2

x =125

Giải: Ta có: \(\dfrac{1}{2.3}\)=\(\dfrac{1}{2}-\dfrac{1}{3}\) ; \(\dfrac{1}{2}-\dfrac{1}{3}\)(giữ nguyên)

Vì \(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{1}{3}\) nên \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

Vậy \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)

\(\dfrac{1}{2}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)

=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+\(\dfrac{1}{4.5}\)+\(\dfrac{1}{5.6}\)+\(\dfrac{1}{6.7}\)+\(\dfrac{1}{7.8}\)+\(\dfrac{1}{8.9}\)+\(\dfrac{1}{9.10}\)

=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+\(\dfrac{1}{9}\)-\(\dfrac{1}{10}\)

=\(\dfrac{1}{1}\)-\(\dfrac{1}{10}\)=\(\dfrac{10}{10}\)-\(\dfrac{1}{10}\)=\(\dfrac{9}{10}\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}=\dfrac{9}{10}\)

P= \(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.17}\)+...+\(\dfrac{3}{96.101}\)

\(\dfrac{5}{3}\).P= \(\dfrac{5}{3}\).(\(\dfrac{3}{1.6}\)+\(\dfrac{3}{6.11}\)+\(\dfrac{3}{11.16}\)+...+\(\dfrac{3}{96.101}\))

\(\dfrac{5}{3}\).P= \(\dfrac{5}{1.6}\)+\(\dfrac{5}{6.11}\)+\(\dfrac{5}{11.16}\)+...+\(\dfrac{5}{96.101}\)

\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{6}\)+\(\dfrac{1}{6}\)-\(\dfrac{1}{11}\)+\(\dfrac{1}{11}\)-\(\dfrac{1}{16}\)+...+\(\dfrac{1}{96}\)-\(\dfrac{1}{101}\)

\(\dfrac{5}{3}\).P= \(\dfrac{1}{1}\)-\(\dfrac{1}{101}\)= \(\dfrac{101}{101}\)-\(\dfrac{1}{101}\)=\(\dfrac{100}{101}\)

P= \(\dfrac{100}{101}\):\(\dfrac{5}{3}\)= \(\dfrac{100}{101}\).\(\dfrac{3}{5}\)=\(\dfrac{100.3}{101.5}\)=\(\dfrac{20.3}{101.1}\)=\(\dfrac{60}{101}\)

Vậy P= \(\dfrac{60}{101}\)

Giải

a)Số học sinh loại giỏi là: 40.\(\dfrac{1}{5}\)= 8 (học sinh)

Số học sinh loại khá là: 8.\(\dfrac{5}{2}\)= 20 (học sinh)

Số học sinh loại trung bình là: 20.50%=10(học sinh)

Số học sinh loại Yếu là: 40-(8+20+10)=2 (học sinh)

b)Tỉ số % của hs yếu so vs cả lớp là: \(\dfrac{2.100}{40}\)% = 5%

Đáp số : a) giỏi : 8 hs ; khá : 20 hs ; trung bình : 10 hs ; yếu : 2 hs

b)5%

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{1}{1}\)-\(\dfrac{1}{100}\)=\(\dfrac{100}{100}\)-\(\dfrac{1}{100}\)=\(\dfrac{99}{100}\)

Vì \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<\(\dfrac{99}{100}\)<1 nên \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<1

Vậy \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{100^2}\)<1.