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2,
\(M=\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\) =\(\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)
\(=\dfrac{3}{4}\)
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
\(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}\)
\(A=\frac{19}{24}+\frac{-1}{2}+\frac{-1}{3}+\frac{-7}{24}\)
\(A=\left(\frac{19}{24}+\frac{-7}{24}\right)+\frac{-1}{2}+\frac{-1}{3}\)
\(A=\frac{1}{2}+\frac{-1}{2}+\frac{-1}{3}\)
\(A=0+\frac{-1}{3}=\frac{-1}{3}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{15}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}+\frac{-3}{7}+\frac{-1}{3}\)
\(B=\left(\frac{49}{168}+\frac{140}{168}+\frac{42}{168}\right)+\left(\frac{-72}{168}+\frac{-56}{168}\right)\)
\(B=\frac{231}{168}+\frac{-128}{168}=\frac{103}{168}\)
Có: \(A=\frac{-1}{3}=\frac{\left(-1\right)\cdot56}{3\cdot56}=\frac{-56}{168}\)
Mặt khác: \(-56< 103\)
\(\Rightarrow\)\(\frac{-56}{168}< \frac{103}{168}\)
\(hay\) \(A< B\)
a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)
b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)
c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)
d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)
c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
a) \(\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{11}{21}\)
\(\Rightarrow x=\dfrac{3\cdot11}{21}\)
\(\Rightarrow x=\dfrac{33}{21}\)
\(\Rightarrow x=\dfrac{11}{7}\)
b) \(\dfrac{x}{5}=\dfrac{5}{6}+\dfrac{-19}{30}\)
\(\Rightarrow\dfrac{x}{5}=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{5\cdot1}{5}\)
\(\Rightarrow x=1\)
a: \(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\)
\(=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{122}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=5\cdot\dfrac{3}{2}=\dfrac{15}{2}\)
b: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}\)
\(=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{35}{6}\cdot\dfrac{30}{-31}-\dfrac{11}{31}\)
\(=\dfrac{-35\cdot5-11}{31}=\dfrac{-186}{31}=-6\)
Tính để mai làm mà thoii kệ, làm luôn :vv
Giải:
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+..+\dfrac{1}{19}\)
\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+..+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}\right)\)
Vì \(\dfrac{1}{5}+\dfrac{1}{6}+..+\dfrac{1}{9}>\dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{9}\)
Nên \(\dfrac{1}{5}+\dfrac{1}{6}+..+\dfrac{1}{9}>\dfrac{5}{9}>\dfrac{1}{2}\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{1}{19}+\dfrac{1}{19}+...+\dfrac{1}{19}\)
Nên \(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{19}>\dfrac{10}{19}>\dfrac{1}{2}\)
\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{1}{2}\)
\(\Rightarrow B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}>1\)
Giải:
Ta có: \(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\)
\(=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{19}\right)\)
Dễ thấy:
\(5< 20\Leftrightarrow\dfrac{1}{5}>\dfrac{1}{20}\)
\(6< 20\Leftrightarrow\dfrac{1}{6}>\dfrac{1}{20}\)
\(....................\)
\(19< 20\Leftrightarrow\dfrac{1}{19}>\dfrac{1}{20}\)
Cộng vế theo vế ta có:
\(B>\dfrac{1}{4}+\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\) (có \(15\) phân số \(\dfrac{1}{20}\))
\(\Rightarrow B>\dfrac{1}{4}+\dfrac{1}{20}.15=\dfrac{1}{4}+\dfrac{3}{4}=1\)
Vậy \(B>1\) (Đpcm)