a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

\(\sqrt{\dfrac{1}{4}+\dfrac{1}{\left(2n-1\right)^2}+\dfrac{1}{\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(2n-1\right)^2\left(2n+1\right)^2+4\left(2n-1\right)^2+4\left(2n+1\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(4n^2-1\right)^2+4\left(4n^2-4n+1\right)+4\left(4n^2+4n+1\right)}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)

\(=\sqrt{\dfrac{16n^4+24n^2+9}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(4n^2+3\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\dfrac{4n^2+3}{2\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{\left(4n^2-1\right)+4}{2\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(=\dfrac{1}{2}+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)

Do đó:

\(P=\left(\dfrac{1}{2}+\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{399}-\dfrac{1}{401}\right)\)

\(=\dfrac{1}{2}.200+1-\dfrac{1}{401}=\dfrac{40500}{401}\)

\(\Rightarrow Q=400\)

Bài 1:

a) 

\(\dfrac{1}{2}=\dfrac{1\times6}{2\times6}=\dfrac{6}{12}\)

\(\dfrac{2}{3}=\dfrac{2\times4}{3\times4}=\dfrac{8}{12}\)

\(\dfrac{3}{4}=\dfrac{3\times3}{4\times3}=\dfrac{9}{12}\)

b)

\(\dfrac{1}{3}=\dfrac{1\times15}{3\times15}=\dfrac{15}{45}\)

\(\dfrac{2}{15}=\dfrac{2\times3}{15\times3}=\dfrac{6}{45}\)

\(\dfrac{4}{45}\) (giữ nguyên)

c)

\(\dfrac{1}{8}=\dfrac{1\times3}{8\times3}=\dfrac{3}{24}\)

\(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)

\(\dfrac{5}{2}=\dfrac{5\times12}{2\times12}=\dfrac{60}{24}\)

d)

\(\dfrac{2}{7}=\dfrac{2\times4}{7\times4}=\dfrac{8}{28}\)

\(\dfrac{9}{4}=\dfrac{9\times7}{4\times7}=\dfrac{63}{28}\)

\(\dfrac{5}{28}\) (giữ nguyên)

Bài 2:

a)

\(4=\dfrac{4}{1}=\dfrac{4\times12}{1\times12}=\dfrac{48}{12}\)

\(\dfrac{9}{4}=\dfrac{9\times3}{4\times3}=\dfrac{27}{12}\)

b)

\(\dfrac{5}{8}=\dfrac{5\times30}{8\times30}=\dfrac{150}{240}\)

\(\dfrac{25}{30}=\dfrac{5}{6}=\dfrac{5\times40}{6\times40}=\dfrac{200}{240}\)

\(2=\dfrac{2}{1}=\dfrac{2\times240}{1\times240}=\dfrac{480}{240}\).

giúp mình với ạ!

a) = =

b) = = = . ( Với điều kiện b # 1)

c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= = = ( với điều kiện a#b).

d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = = = =

a) Các phân số đảo ngược là:

\(\dfrac{5}{8}\rightarrow\dfrac{8}{5};\dfrac{3}{4}\rightarrow\dfrac{4}{3};\dfrac{1}{2}\rightarrow\dfrac{2}{1}=2\)

b) \(\dfrac{3}{7}:\dfrac{5}{8}=\dfrac{3}{7}\times\dfrac{8}{5}=\dfrac{24}{35}\)

\(\dfrac{8}{7}:\dfrac{3}{4}=\dfrac{8}{7}\times\dfrac{4}{3}=\dfrac{32}{21}\)

\(\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{1}{3}\times2=\dfrac{2\times1}{3}=\dfrac{2}{3}\)

a, 5/2 + 2/3 - 3/4

= 19/6 - 3/4

= 29/12

b, 4/5 - 1/2 + 1/3

= 3/10 + 1/3

= 19/30

c, 2/5 x 1/2 : 1/3

= 1/5 : 1/3

= 3/5

\(a.\)

\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)

\(\Leftrightarrow?=-3\)

\(b.\)

\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)

\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)

\(\Leftrightarrow?=-2\)

\(c.\)

\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow?=10\)

Mk gọi ? = x nha

a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\) 

           \(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\) 

           \(\dfrac{x}{4}=\dfrac{-3}{4}\) 

⇒x=-3

b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)  

       \(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\) 

       \(\dfrac{x}{3}=\dfrac{-2}{3}\) 

⇒x=-2

c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\) 

        \(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\) 

        \(\dfrac{3}{x}=\dfrac{3}{10}\) 

⇒x=10

bài này đúng là thị của phi...vô của lí ... :))