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= \(49-\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=\dfrac{4949}{102}\notin N\)
Vậy \(S\notin N\)
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)\(S=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{2499}{2500}\)
\(\Rightarrow S=1-\dfrac{1}{4}+1-\dfrac{1}{9}+1-\dfrac{1}{16}+...+1-\dfrac{1}{2500}\)
\(\Rightarrow S=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+1-\dfrac{1}{4^2}+...+1-\dfrac{1}{50^2}\)
\(\Rightarrow S=\left(1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)
Từ 2-50 có 49 số nên có 49 số 1
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)< 49\)
Nhận xét: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{50^2}< \dfrac{1}{49.50}\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{50}=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>49-1\)
\(\Rightarrow S=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)>48\) (2)
Từ (1) và (2) \(\Rightarrow48< S< 49\)
Vậy \(S\notin N\)
\(B=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}\)
\(=1-\dfrac{3}{4}+1-\dfrac{8}{9}+1-\dfrac{15}{16}+1-\dfrac{24}{25}...+1-\dfrac{2499}{2500}\)
\(=49-\left(\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+...+\dfrac{1}{2500}\right)\)
Lại có: \(49-\left(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+\dfrac{1}{5.5}+...+\dfrac{1}{50.50}\right)< 49-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\right)\)
Mà \(49-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{50.51}\right)\)
\(=49-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(=49-\left(\dfrac{1}{2}-\dfrac{1}{51}\right)=\dfrac{4942}{102}\) \(\notin Z\)
Vậy B không phải là số nguyên
a, \(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)=\dfrac{2}{3}-4.\dfrac{5}{4}=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
b, \(-\dfrac{2}{3}.\dfrac{3}{11}+\dfrac{-16}{9}.\dfrac{3}{11}=\dfrac{3}{11}.\left(-\dfrac{2}{3}+\dfrac{-16}{9}\right)\)
\(=\dfrac{3}{11}.\dfrac{-22}{9}=-\dfrac{2}{3}\)
Câu c lấy máy tính tính nha!
c,
c) \(\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{5}{7}+\dfrac{4}{9}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}-\dfrac{17}{18}-\dfrac{10}{14}+\dfrac{8}{18}+\dfrac{17}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{17}{18}+\dfrac{8}{18}\right)+\left(-\dfrac{10}{14}+\dfrac{17}{14}\right)\)
\(=\dfrac{11}{125}+\left(-\dfrac{9}{18}\right)+\dfrac{7}{14}\)
\(=\dfrac{11}{125}+\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}\)
\(=\dfrac{11}{125}+0=\dfrac{11}{125}\)
a.\(\dfrac{17}{15}\div\dfrac{4}{3}=\dfrac{17}{20}\)
b.\(\dfrac{-12}{21}\div\dfrac{34}{43}=\dfrac{-86}{119}\)
c.\(\dfrac{-5}{9}\times\dfrac{3}{11}+\dfrac{13}{18}\times\dfrac{3}{11}\)
=\(\dfrac{3}{11}\times(\dfrac{-5}{9}+\dfrac{13}{18})=\dfrac{3}{11}\times\dfrac{1}{6}=\dfrac{1}{22}\)
d.\(\dfrac{-2}{9}\times\dfrac{5}{11}+\dfrac{-16}{9}\times\dfrac{5}{11}=\dfrac{5}{11}\times(\dfrac{-2}{9}+\dfrac{-16}{9})\)
=\(\dfrac{5}{11}\times(-2)=\dfrac{-10}{11}\)
b: \(\left(\dfrac{2}{5}-\dfrac{7}{10}x\right):\dfrac{5}{3}=-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{2}{5}-\dfrac{7}{10}x=\dfrac{-3}{4}\cdot\dfrac{5}{3}=\dfrac{-5}{4}\)
\(\Leftrightarrow x\cdot\dfrac{7}{10}=\dfrac{2}{5}+\dfrac{5}{4}=\dfrac{8+25}{20}=\dfrac{33}{20}\)
\(\Leftrightarrow x=\dfrac{33}{20}:\dfrac{7}{10}=\dfrac{33}{20}\cdot\dfrac{10}{7}=\dfrac{33}{14}\)
c: \(\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)-\dfrac{11}{6}=0\)
\(\Leftrightarrow\dfrac{7}{16}:\left(\dfrac{1}{4}x+\dfrac{9}{2}\right)=\dfrac{11}{6}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}+\dfrac{9}{2}=\dfrac{11}{6}:\dfrac{7}{16}=\dfrac{88}{21}\)
\(\Leftrightarrow x\cdot\dfrac{1}{4}=\dfrac{88}{21}-\dfrac{9}{2}=-\dfrac{13}{42}\)
hay \(x=-\dfrac{26}{21}\)
a. = 1/20 + 5 - 1/2
= 101/20 - 1/2
= 91/20
b. = ( 6/15 - 3/5) - ( 7/8 + 2/16) + 3
= -1/5 - 1 + 3
= 9/5
c. = 15/7 . ( 3/5 - 8/5)
= 15/7 . ( -1)
= - 15/7
e. = -14/9 - 3/9
= -17/9
f. = 19/21 . ( 15/17 + 2/17) + 13/21
= 19/21 . 1 + 13/21
= 32/21
g. = 43/12 : 2 + 5/24
= 43/24 + 5/24
= 2