Các dạng toán về hỗn số dương

$- 2\dfrac{3}{4} =  - \dfrac{{2.4 + 3}}{4} =  - \dfrac{{11}}{4}$

Đáp án A:

$\dfrac{1}{{19}} + \dfrac{1}{{20}} = \dfrac{{20}}{{19.20}} + \dfrac{{19}}{{19.20}}$ $= \dfrac{{19 + 20}}{{19.20}} \ne \dfrac{{19.20}}{{19 + 20}}$

Nên A sai.

Đáp án B: $6\dfrac{{23}}{{11}} = \dfrac{{6.11 + 23}}{{11}} \ne \dfrac{{6.23 + 11}}{{11}}$ nên B sai.

Đáp án C: $a\dfrac{a}{{99}} = \dfrac{{a.99 + a}}{{99}}$$= \dfrac{{a.\left( {99 + 1} \right)}}{{99}} = \dfrac{{100a}}{{99}}$ nên C đúng.

Đáp án D: $1\dfrac{{15}}{{23}} = \dfrac{{1.23 + 15}}{{15}} \ne \dfrac{{1.23}}{{15}}$ nên D sai.

Ta có: $23\%  = \dfrac{{23}}{{100}};$ $- 1\dfrac{1}{{12}} =  - \dfrac{{13}}{{12}};$ $5\dfrac{1}{2} = \dfrac{{11}}{2}$

Ta chia thành hai nhóm phân số là: $\dfrac{{23}}{{100}};\dfrac{{12}}{{100}};\dfrac{{11}}{2}$ và $- \dfrac{{13}}{{12}}; - \dfrac{{31}}{{24}}$

Nhóm 1:

$\dfrac{{12}}{{100}} < \dfrac{{23}}{{100}} < 1 < \dfrac{{11}}{2}$ nên $\dfrac{{12}}{{100}} < \dfrac{{23}}{{100}} < \dfrac{{11}}{2}$

Nhóm 2: $- \dfrac{{13}}{{12}}; - \dfrac{{31}}{{24}}$

$- \dfrac{{13}}{{12}} = \dfrac{{ - 26}}{{24}} > \dfrac{{ - 31}}{{24}}$ nên $- \dfrac{{13}}{{12}} >  - \dfrac{{31}}{{24}}$

Vậy $- \dfrac{{31}}{{24}} <  - \dfrac{{13}}{{12}} < \dfrac{{12}}{{100}} < \dfrac{{23}}{{100}} < \dfrac{{11}}{2}$ hay $- \dfrac{{31}}{{24}} <  - 1\dfrac{1}{{12}} < \dfrac{{12}}{{100}} < 23\%  < 5\dfrac{1}{2}$

$\left( { - 1\dfrac{1}{3}} \right) + 2\dfrac{1}{2} =  - \dfrac{4}{3} + \dfrac{5}{2}$$= \dfrac{{ - 8}}{6} + \dfrac{{15}}{6} = \dfrac{7}{6}$

$\begin{array}{l}x - 3\dfrac{1}{2}x =  - \dfrac{{20}}{7}\\x - \dfrac{7}{2}x =  - \dfrac{{20}}{7}\\x.\left( {1 - \dfrac{7}{2}} \right) =  - \dfrac{{20}}{7}\\x.\left( {\dfrac{{ - 5}}{2}} \right) = \dfrac{{ - 20}}{7}\\x = \dfrac{{ - 20}}{7}:\dfrac{{ - 5}}{2}\\x = \dfrac{{ - 20}}{7}.\dfrac{2}{{ - 5}}\\x = \dfrac{8}{7} \\x= 1\dfrac{1}{7}\end{array}$

$A = \left( {4\dfrac{5}{{17}} - 3\dfrac{4}{5} + 8\dfrac{{15}}{{29}}} \right) - \left( {3\dfrac{5}{{17}} - 6\dfrac{{14}}{{29}}} \right)$

$A = 4\dfrac{5}{{17}} - 3\dfrac{4}{5} + 8\dfrac{{15}}{{29}} - 3\dfrac{5}{{17}} + 6\dfrac{{14}}{{29}}$

$A = \left( {4\dfrac{5}{{17}} - 3\dfrac{5}{{17}}} \right) + \left( {8\dfrac{{15}}{{29}} + 6\dfrac{{14}}{{29}}} \right) - 3\dfrac{4}{5}$

$A = \left( {4 - 3} \right) + \left( {\dfrac{5}{{17}} - \dfrac{5}{{17}}} \right)$ $+ \left( {8 + 6} \right) + \left( {\dfrac{{15}}{{29}} + \dfrac{{14}}{{29}}} \right) - 3\dfrac{4}{5}$

$A = 1 + 0 + 14 + 1 - 3\dfrac{4}{5}$

$A=16-3\dfrac{4}{5}$

$A = 15\dfrac{5}{5} - 3\dfrac{4}{5} = 12\dfrac{1}{5}$

$N =  - \dfrac{1}{7}\left( {9\dfrac{1}{2} - 8,75} \right):\dfrac{2}{7} + 0,625:1\dfrac{2}{3}$

$N =  - \dfrac{1}{7}\left( {\dfrac{{19}}{2} - \dfrac{{875}}{{100}}} \right).\dfrac{7}{2} + \dfrac{{625}}{{1000}}:\dfrac{5}{3}$

$N =  - \dfrac{1}{7}\left( {\dfrac{{19}}{2} - \dfrac{{35}}{4}} \right).\dfrac{7}{2} + \dfrac{5}{8}.\dfrac{3}{5}$

$N =  - \dfrac{1}{7}.\dfrac{3}{4}.\dfrac{7}{2} + \dfrac{3}{8}$

$N =  - \dfrac{3}{8} + \dfrac{3}{8} = 0$

$\dfrac{{\left( {1,16 - x} \right).5,25}}{{\left( {10\dfrac{5}{9} - 7\dfrac{1}{4}} \right).2\dfrac{2}{{17}}}} = 75\%$

$\dfrac{{\left( {\dfrac{{116}}{{100}} - x} \right).\dfrac{{525}}{{100}}}}{{\left( {\dfrac{{95}}{9} - \dfrac{{29}}{4}} \right).\dfrac{{36}}{{17}}}} = \dfrac{{75}}{{100}}$

$\dfrac{{\left( {\dfrac{{29}}{{25}} - x} \right).\dfrac{{21}}{4}}}{{\dfrac{{119}}{{36}}.\dfrac{{36}}{{17}}}} = \dfrac{3}{4}$

$\dfrac{{\left( {\dfrac{{29}}{{25}} - x} \right).\dfrac{{21}}{4}}}{7} = \dfrac{3}{4}$

$\left( {\dfrac{{29}}{{25}} - x} \right).\dfrac{{21}}{4}.4 = 7.3$

$\left( {\dfrac{{29}}{{25}} - x} \right).21 = 21$

$\dfrac{{29}}{{25}} - x = 21:21$

$\dfrac{{29}}{{25}} - x = 1$

$x = \dfrac{{29}}{{25}} - 1$

$x = \dfrac{4}{{25}}$

Ta có:

$3\dfrac{3}{4}$ tạ = $\dfrac{{15}}{4}$ tạ = $\dfrac{{375}}{{100}}$ tạ.

$\dfrac{7}{2}$ tạ = $\dfrac{{350}}{{100}}$ tạ

$3\dfrac{{45}}{{100}}$ tạ = $\dfrac{{345}}{{100}}$ tạ

$365$kg = $\dfrac{{365}}{{100}}$ tạ

=> Các khối lượng theo thứ tự từ lớn đến nhỏ là:

$\dfrac{{377}}{{100}}$ tạ ; $3\dfrac{3}{4}$ tạ; $365$kg; $\dfrac{7}{2}$ tạ; $3\dfrac{{45}}{{100}}$ tạ.

a) $125\,d{m^2} = \dfrac{{125}}{{100}}{m^2} = 1\dfrac{{25}}{{100}}\,{m^2}$

b) $218\,c{m^2} = \dfrac{{218}}{{10000}}{m^2} = \dfrac{{109}}{{5000}}\,{m^2}$

c) $240\,d{m^2} = \dfrac{{240}}{{100}}{m^2} = 2\dfrac{{40}}{{100}}\,{m^2}$

d) $34\,c{m^2} = \dfrac{{34}}{{10000}}{m^2} = \dfrac{{17}}{{5000}}\,{m^2}$

Vậy ta được: $1\dfrac{{25}}{{100}}\,{m^2}$; $\dfrac{{109}}{{5000}}\,{m^2}$; $2\dfrac{{40}}{{100}}\,{m^2}$; $\dfrac{{17}}{{5000}}\,{m^2}$.

Đổi 70 phút = $\dfrac{7}{6}$ giờ

Vận tốc của xe taxi là:

100 : $1\dfrac{1}{5}$ = 100 : $\dfrac{6}{5}$ = $\dfrac{{250}}{3}$ = $83\dfrac{1}{3}$ (km/h)

Vận tốc của xe tải là:

100 : $\dfrac{7}{6}$ = $\dfrac{{600}}{7}$ = $85\dfrac{5}{7}$ (km/h)

Ta có: $85\dfrac{5}{7}$ > $83\dfrac{1}{3}$ nên vận tốc của xe tải lớn hơn vận tốc xe taxi.