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Chỗ phức tạp là ở biểu thức trong ngoặc thôi
Ta có
\(\dfrac{1}{8}+\dfrac{1}{8\cdot15}+\dfrac{1}{15\cdot22}...+\dfrac{1}{43\cdot50}\)
\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{22}+....+\dfrac{1}{43}-\dfrac{1}{50}\right)\right]\)
\(=\dfrac{1}{8}\cdot\left[\dfrac{1}{7}\left(\dfrac{1}{8}-\dfrac{1}{50}\right)\right]=\dfrac{1}{8}\cdot\dfrac{3}{200}=\dfrac{3}{1600}\)
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 \(B=\frac{17}{7}\Leftrightarrow x=0\)
đặt A = 1/4.9 + 1/9.14+ 1/14.19 + .....1/44.49
ta có 5.A = 5/4.9 + 5/9.14+ 5/14.19 + .....5/44.49 = 1/4- 1//9 + 1/9 - 1/14+........+ 1/44 -1/49 = 1/4 - 149 = 45/196
suy ra A = 9/196
đặt B = 1-3--5-...-49 = 1 - (3+5+ ....+ 49)
đặt C = 3+5+...+49 khoảng cách là d = 2
số các số hạng là (49-3)/2 + 1 = 24
tổng C = (49+3)/2 x 24 = 624
suy ra B = 1-624 = -623
vậy A = 9/196 .(-623)/89 = -9/28
\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left[\left(\frac{2}{3}-\frac{4}{7}\right)+\left(-\frac{8}{7}+\frac{1}{3}\right)\right]:\frac{5}{9}\)
\(=\left(\frac{2}{3}-\frac{4}{7}-\frac{8}{7}+\frac{1}{3}\right)\cdot\frac{9}{5}\)
\(=\left(1-\frac{12}{7}\right)\cdot\frac{9}{5}\)
\(=-\frac{5}{7}\cdot\frac{9}{5}\)
\(-\frac{9}{7}\)
\(\left(\frac{2}{3}-\frac{4}{7}\right):\frac{5}{9}+\left(-\frac{8}{7}+\frac{1}{3}\right):\frac{5}{9}\)
\(=\left(\frac{14}{21}-\frac{12}{21}\right):\frac{5}{9}+\left(-\frac{24}{21}+\frac{7}{21}\right):\frac{5}{9}\)
\(=\frac{2}{21}:\frac{5}{9}+\frac{-17}{21}:\frac{5}{9}\)
\(=\left(\frac{2}{21}+\frac{-17}{21}\right):\frac{5}{9}\)
\(=\frac{-15}{21}:\frac{5}{9}\)
\(=\frac{-15}{21}.\frac{9}{5}\)
\(=\frac{-9}{7}\)
\(=\frac{99}{35}\)