so sánh : \(\frac{-2}{5}\), \(\frac{-4}{9}\), \(\frac{6}{18}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~
1.
\(a.\frac{1}{2}+a.\frac{1}{4}=-\frac{4}{5}\Rightarrow a.\left(\frac{1}{2}+\frac{1}{4}\right)=-\frac{4}{5}\Rightarrow a=-\frac{16}{15}\)
2. Ta có:
\(A=\left(2\frac{5}{6}+1\frac{4}{9}\right):\left(10\frac{1}{12}-9\frac{1}{2}\right)=\left(\frac{17}{6}+\frac{13}{9}\right):\left(\frac{121}{12}-\frac{19}{2}\right)=\frac{77}{18}:\frac{7}{12}=\frac{22}{3}\)
\(B=1\frac{5}{18}-\frac{5}{18}\left(\frac{1}{15}+1\frac{1}{3}\right)=\frac{23}{18}-\frac{5}{18}\left(\frac{1}{15}+\frac{4}{3}\right)=\frac{23}{18}-\frac{1}{54}-\frac{10}{27}=\frac{8}{9}\)
Có: \(\frac{22}{3}=\frac{66}{9}>\frac{8}{9}\Leftrightarrow A>B\)
Ta có
\(C=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}...+\frac{1}{17.18}>A=\frac{1}{2.3}+\frac{1}{5.4}+...+\frac{1}{18.19}\)
\(C< =>\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{18-17}{17.18}\)\(>A\)
\(C< =>\frac{1}{2}-\frac{1}{18}\)\(>A\)
\(C< =>\frac{4}{9}\)\(>A\left(1\right)\)
Lại có \(C=\frac{4}{9}< \frac{9}{19}=B\left(2\right)\)
Từ (1),(2) => B>A
Ta có: \(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{1}{5}.4=\frac{4}{5}\)
\(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}< \frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{1}{10}.4=\frac{2}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}< \frac{1}{15}+\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\frac{1}{15}.4=\frac{4}{15}\)
\(\frac{1}{17}+\frac{1}{18}< \frac{1}{17}.2=\frac{2}{17}\)
\(\Rightarrow B< \frac{4}{5}+\frac{2}{5}+\frac{4}{15}+\frac{2}{17}=\frac{404}{255}=1\frac{149}{255}< 2\)
a) $\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$
Ta có $\frac{{12}}{{18}} > \frac{{11}}{{18}}$ nên $\frac{2}{3} > \frac{{11}}{{18}}$
b) $\frac{{36}}{{63}} = \frac{{36:9}}{{63:9}} = \frac{4}{7}$
Ta có $\frac{4}{7} < \frac{5}{7}$ nên $\frac{{36}}{{63}}$ < $\frac{5}{7}$
c)
$\frac{{55}}{{110}} = \frac{{55:55}}{{110:55}} = \frac{1}{2}$ ; $\frac{4}{8} = \frac{1}{2}$
Vậy $\frac{{55}}{{110}}$ = $\frac{4}{8}$