Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1 :
\(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\frac{4-3-5-7-...-49}{217}\)
\(=\frac{1}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right).\frac{5-\left(1+3+5+7+...+49\right)}{217}\)
\(=\frac{1}{7}\left(1-\frac{1}{50}\right).\frac{5-\left(12.50\right)+25}{217}\)
\(=\frac{1}{7}.\frac{49}{50}.\frac{5-625}{217}\)
\(=\frac{-2}{5}\)
Bài 2 :
\(B=\frac{x^2+17}{x^2+7}=\frac{\left(x^2+7\right)+10}{x^2+7}=1+\frac{10}{x^2+7}\)
Ta có : \(x^2\ge0\). Dấu '' = '' xảy ra khi :
\(x=0\Rightarrow x^2+7\ge7\)( 2 vế dương )
\(\Rightarrow\frac{10}{x^2+7}\le\frac{10}{7}\)
\(\Rightarrow1+\frac{10}{x^2+7}\le1+\frac{10}{7}\)
\(\Rightarrow B\le\frac{17}{7}\)
Dấu '' = '' xảy ra < = > x = 0
Vậy Max \(B=\frac{17}{7}\Leftrightarrow x=0\)
Chỗ phức tạp là ở biểu thức trong ngoặc thôi
Ta có
\(\dfrac{1}{8}+\dfrac{1}{8\cdot15}+\dfrac{1}{15\cdot22}...+\dfrac{1}{43\cdot50}\)
\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{22}+....+\dfrac{1}{43}-\dfrac{1}{50}\right)\right]\)
\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{50}\right)\right]=\dfrac{1}{8}\cdot\dfrac{3}{200}=\dfrac{3}{1600}\)
A = \(\left(\frac{1}{8}+\frac{1}{8.15}+\frac{1}{15.22}+...+\frac{1}{43.50}\right)\cdot\frac{4-3-5-7-...-49}{217}\)
A = \(\frac{1}{7}.\left(\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{43.50}\right)\cdot\frac{4-\left(3+5+7+...+49\right)}{217}\)
A = \(\frac{1}{7}.\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+...+\frac{1}{43}-\frac{1}{50}\right)\cdot\frac{4-\left(49+3\right)\left[\left(49-3\right):2+1\right]:2}{217}\)
A = \(\frac{1}{7}\cdot\left(1-\frac{1}{50}\right)\cdot\frac{4-52.24:2}{217}\)
A = \(\frac{1}{7}\cdot\frac{49}{50}\cdot\frac{4-624}{217}\)
A = \(\frac{7}{50}\cdot\frac{-620}{217}=-\frac{2}{5}\)
\(A=\frac{3}{1.8}+\frac{3}{8.15}+\frac{3}{15.22}+...+\frac{3}{106.113}\)
\(=\frac{3}{7}\left(\frac{7}{1.8}+\frac{7}{8.15}+\frac{7}{15.22}+...+\frac{7}{106.113}\right)\)
\(=\frac{3}{7}\left(\frac{8-1}{1.8}+\frac{15-8}{8.15}+\frac{22-15}{15.22}+...+\frac{113-106}{106.113}\right)\)
\(=\frac{3}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+\frac{1}{106}-\frac{1}{113}\right)\)
\(=\frac{3}{7}\left(1-\frac{1}{113}\right)=\frac{48}{113}\)
\(B=\frac{25}{50.55}+\frac{25}{55.60}+...+\frac{25}{95.100}\)
\(=5\left(\frac{5}{50.55}+\frac{5}{55.60}+...+\frac{5}{95.100}\right)\)
\(=5\left(\frac{55-50}{50.55}+\frac{60-55}{55.60}+...+\frac{100-95}{95.100}\right)\)
\(=5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=5\left(\frac{1}{50}-\frac{1}{100}\right)=\frac{1}{20}\)
Giá trị của biểu thức đã cho là:
\(A-B=\frac{48}{113}-\frac{1}{20}=\frac{847}{2260}\)