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a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)
Bạn tham khảo nhé
Ta có :
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}\)
\(B=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+\frac{5^2-1}{5^2}+...+\frac{50^2-1}{50^2}\)
\(B=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{4^2}\right)+\left(1-\frac{1}{5^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)
\(B=\left(1+1+1+1+...+1\right)-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-\frac{1}{5^2}-...-\frac{1}{50^2}\)
\(B=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\right)\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< 1-\frac{1}{50}\)
\(A< \frac{49}{50}\)\(\left(1\right)\)
Lại có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{50^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{50.51}\)
\(A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{50}-\frac{1}{51}\)
\(A>\frac{1}{2}-\frac{1}{51}=\frac{49}{102}\)\(\left(2\right)\)
Từ (1) và (2) suy ra \(\frac{49}{102}< A< \frac{49}{50}\)
\(\Leftrightarrow\)\(49-\frac{49}{102}< 49-A< 49-\frac{49}{50}\)
\(\Leftrightarrow\)\(\frac{4949}{102}< B< \frac{2401}{50}\)
\(\Rightarrow\)\(B\notinℤ\)
Vậy B không là số nguyên
\(=\frac{2\cdot4}{3^2}\cdot\frac{3.5}{4^2}\cdot\frac{4\cdot6}{5^2}\cdot......\cdot\frac{49\cdot51}{50^2}\)
=\(\frac{\left[2\cdot3\cdot4\cdot......\cdot49\right]\cdot\left[4\cdot5\cdot6\cdot.....\cdot51\right]}{\left[3\cdot4\cdot5\cdot....\cdot50\right]\cdot\left[3\cdot4\cdot5\cdot....\cdot50\right]}\)
=\(\frac{2\cdot51}{50\cdot3}\)
=\(\frac{17}{25}\)
Vì \(\frac{17}{25}\) ko phải là số nguyên nên B ko phải là số nguyên [ĐPCM]
Bài 2:
1) \(\frac{x}{12}-\frac{5}{6}=\frac{1}{12}\)
\(\Rightarrow\frac{x}{12}=\frac{1}{12}+\frac{5}{6}\)
\(\Rightarrow\frac{x}{12}=\frac{11}{12}\)
\(\Rightarrow x.12=11.12\)
\(\Rightarrow x.12=132\)
\(\Rightarrow x=132:12\)
\(\Rightarrow x=11\)
Vậy \(x=11.\)
2) \(\frac{2}{3}-1\frac{4}{15}x=\frac{-3}{5}\)
\(\Rightarrow\frac{2}{3}-\frac{19}{15}x=\frac{-3}{5}\)
\(\Rightarrow\frac{19}{15}x=\frac{2}{3}+\frac{3}{5}\)
\(\Rightarrow\frac{19}{15}x=\frac{19}{15}\)
\(\Rightarrow x=\frac{19}{15}:\frac{19}{15}\)
\(\Rightarrow x=1\)
Vậy \(x=1.\)
3) \(\left(-2\right)^3+0,5x=1,5\)
\(\Rightarrow-8+0,5x=1,5\)
\(\Rightarrow0,5x=1,5+8\)
\(\Rightarrow0,5x=9,5\)
\(\Rightarrow x=9,5:0,5\)
\(\Rightarrow x=19\)
Vậy \(x=19.\)
Chúc bạn học tốt!
