a)

\(\dfrac{5}{8}\left(kg\right)< \dfrac{6}{8}\left(kg\right)\)

Do 13 > 12 nên \(\dfrac{13}{12}>1\)

hay: \(\dfrac{13}{12}\left(kg\right)>1\left(kg\right)\)

b)

\(\dfrac{11}{12}\left(l\right)>\dfrac{11}{14}\left(l\right)\)

Có:

\(\dfrac{5}{3}=\dfrac{5\times3}{3\times3}=\dfrac{15}{9}\)

nên: \(\dfrac{5}{3}\left(l\right)=\dfrac{15}{9}\left(l\right)\)

c)

\(\dfrac{5}{6}=\dfrac{5\times3}{6\times3}=\dfrac{15}{18n}\)

nên \(\dfrac{5}{6}\left(m\right)< \dfrac{17}{8}\left(m\right)\)

Có: \(2=\dfrac{2\times7}{7}=\dfrac{14}{7}\)

nên: \(\dfrac{16}{7}\left(m\right)>2\left(m\right)\)

a, <

b,>

c,<

d,>

e,=

g,>

\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)

\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)

\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)

\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)

\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)

Vậy \(x\in\left\{22;23;24;...\right\}\)

\(\dfrac{????????}{????????????}\)

a) \(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)

   \(\dfrac{11}{55}+\dfrac{10}{55}< \dfrac{x}{55}< \dfrac{22}{55}+\dfrac{1}{55}\)

   \(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{23}{55}\)

\(\Rightarrow\) \(x=22\)

b) \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}< x\le\dfrac{13}{4}+\dfrac{14}{8}\)

  \(\dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6}< x\le\dfrac{26}{8}+\dfrac{14}{8}\)

  \(1< x\le5\)

  \(\Rightarrow\) \(x\in\) {\(2;3;4;5\)}

c) \(\dfrac{1}{3}+\dfrac{5}{12}+\dfrac{-1}{13}< x< \dfrac{7}{5}+\dfrac{2}{10}+\dfrac{1}{2}\)

 Ko biết làm

d) \(\dfrac{79}{15}+\dfrac{7}{5}+\dfrac{-8}{3}\le x\le\dfrac{10}{3}+\dfrac{15}{4}+\dfrac{23}{12}\)

   \(\dfrac{79}{15}+\dfrac{21}{15}+\dfrac{-40}{15}\le x\le\dfrac{40}{12}+\dfrac{45}{12}+\dfrac{23}{12}\)

   \(4\le x\le9\)

   \(\Rightarrow\) \(x\in\) {\(4;5;6;7;8;9\)}

cíu tui trời ơi

b: =12+5/14-3-5/7-5-5/14

=4-5/7

=28/7-5/7=23/7

c: =(-2/5-11/10)+(7/11-7/11)

=-4/10-11/10=-15/10=-3/2

\(a,\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{5}{13}\cdot\dfrac{2}{9}\)

\(=\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{2}{13}\cdot\dfrac{5}{9}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}+\dfrac{8}{13}-\dfrac{2}{13}\right)\)

\(=\dfrac{5}{9}\cdot\dfrac{14}{13}\)

\(=\dfrac{70}{117}\)

\(d,\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}+\dfrac{-2}{5}\)

\(=\left(\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}\right)+\dfrac{-2}{5}\)

\(=0+\dfrac{-2}{5}\)

\(=\dfrac{-2}{5}\)

\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\\ =\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\\ =\dfrac{1}{2}+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)+\dfrac{4}{5}\\ =\dfrac{1}{2}+0+\dfrac{4}{5}\\ =\dfrac{1}{2}+\dfrac{4}{5}\\ =\dfrac{5}{10}+\dfrac{8}{10}\\ =\dfrac{13}{10}\)

\(\dfrac{-3}{7}+\dfrac{3}{4}:\dfrac{3}{14}\\ =\dfrac{-3}{7}+\dfrac{3}{4}\cdot\dfrac{14}{3}\\ =\dfrac{-3}{7}+\dfrac{7}{2}\\ =\dfrac{-6}{14}+\dfrac{49}{14}\\ =\dfrac{43}{14}\)

\(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\\ =\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right):\dfrac{7}{3}\\ =\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{5}{20}\right):\dfrac{7}{3}\\ =\dfrac{7}{5}-\dfrac{4}{5}:\dfrac{7}{3}\\ =\dfrac{7}{5}-\dfrac{4}{5}.\dfrac{3}{7}\\ =\dfrac{7}{5}-\dfrac{12}{35}\\ =\dfrac{49}{35}-\dfrac{12}{35}\\ =\dfrac{37}{35}\)

Lời giải:

Ta có:

\(\left\{\begin{matrix} \frac{1}{13}< \frac{1}{12}\\ \frac{1}{14}< \frac{1}{12}\\ \frac{1}{15}< \frac{1}{12}\end{matrix}\right.\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}(1)\)

\(\left\{\begin{matrix} \frac{1}{61}< \frac{1}{60}\\ \frac{1}{62}< \frac{1}{60}\\ \frac{1}{63}< \frac{1}{60}\end{matrix}\right.\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}(2)\)

Từ \((1);(2)\Rightarrow \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)

Hay \( \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)

Ta có đpcm.

Đặt A là biểu thức đó

Ta có:

\(\dfrac{1}{13}< \dfrac{1}{12};\dfrac{1}{14}< \dfrac{1}{12};\dfrac{1}{15}< \dfrac{1}{12}\)

\(\Rightarrow\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}\)

Ta cũng có

\(\dfrac{1}{61}< \dfrac{1}{60};\dfrac{1}{62}< \dfrac{1}{60};\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)

\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

\(\Rightarrow\)dpcm

11: \(=\dfrac{-5}{7}+\dfrac{5}{67}+\dfrac{13}{30}+\dfrac{1}{2}-\dfrac{11}{6}+\dfrac{17}{14}+\dfrac{2}{5}\)

\(=\left(\dfrac{-5}{7}+\dfrac{1}{2}+\dfrac{17}{14}\right)+\left(\dfrac{13}{30}-\dfrac{11}{6}+\dfrac{2}{5}\right)+\dfrac{5}{67}\)

\(=\dfrac{-10+7+17}{14}+\dfrac{13-55+12}{30}+\dfrac{5}{67}\)

\(=1-1+\dfrac{5}{67}=\dfrac{5}{67}\)

12: \(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)

\(=\dfrac{-1}{4}\left(\dfrac{152}{11}+\dfrac{68}{11}\right)=-\dfrac{1}{4}\cdot20=-5\)

a) Giải:

Ta có: \(4n-5=4\left(n-3\right)+7\)

Để \(\left(4n-5\right)⋮\left(n-3\right)\Leftrightarrow7⋮n-3\)

\(\Rightarrow n-3\inƯ\left(7\right)\)

Mà \(Ư\left(7\right)\in\left\{\pm1;\pm7\right\}\)

Nên ta có bảng sau:

Vậy \(n=\left\{2;4;-4;10\right\}\)

b) Ta có:

\(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

\(=\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Nhận xét:

\(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}+\dfrac{1}{12}+\dfrac{1}{12}=\dfrac{1}{4}\)

\(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}+\dfrac{1}{60}+\dfrac{1}{60}=\dfrac{1}{20}\)

\(\Rightarrow S< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)

Vậy \(S=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) \(< \dfrac{1}{2}\) (Đpcm)