Hỗn số dương

Ta có: \(4:3\) bằng $1$ (dư \(1\) ) nên \(\dfrac{4}{3} = 1\dfrac{1}{3}\)

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

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.

\(\left( { - 2\dfrac{1}{4}} \right) + \dfrac{5}{2} =  - \dfrac{9}{4} + \dfrac{5}{2}\)\( = \dfrac{{ - 9}}{4} + \dfrac{{10}}{4} = \dfrac{1}{4}\)

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

\(\begin{array}{l}2\dfrac{x}{7} = \dfrac{{75}}{{35}}\\\dfrac{{2.7 + x}}{7} = \dfrac{{15}}{7}\\14 + x = 15\\x = 15 - 14\\x = 1\end{array}\)

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

Đáp án A: \(\left( { - 3\dfrac{3}{4}} \right).1\dfrac{1}{2}\)\( =  - \dfrac{{15}}{4}.\dfrac{3}{2} =  - \dfrac{{45}}{8} =  - 5\dfrac{5}{8} \ne  - 3\dfrac{3}{8}\)

Nên A sai.

Đáp án B: \(3\dfrac{3}{4}:1\dfrac{1}{5} = \dfrac{{15}}{4}:\dfrac{6}{5} = \dfrac{{15}}{4}.\dfrac{5}{6}\)\( = \dfrac{{25}}{8} = 3\dfrac{1}{8} \ne 3\dfrac{3}{{20}}\) nên B sai.

Đáp án C: \(\left( { - 3} \right) - \left( { - 2\dfrac{2}{5}} \right)\)\( = \left( { - 3} \right) - \left( { - \dfrac{{12}}{5}} \right) = \left( { - 3} \right) + \dfrac{{12}}{5} = \dfrac{{ - 3}}{5}\)

Nên C đúng.

Đáp án D: \(5\dfrac{7}{{10}}.15 = \dfrac{{57}}{{10}}.15 = \dfrac{{171}}{2} \ne \dfrac{{105}}{2}\) nên D sai.

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

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

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

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

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

\(A=16-3\dfrac{4}{5}\)

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

\(M = 60\dfrac{7}{{13}}.x + 50\dfrac{8}{{13}}.x - 11\dfrac{2}{{13}}.x\)

\(M = \left( {60\dfrac{7}{{13}} + 50\dfrac{8}{{13}} - 11\dfrac{2}{{13}}} \right).x\)

\(M = \left[ {\left( {60 + 50 - 11} \right) + \left( {\dfrac{7}{{13}} + \dfrac{8}{{13}} - \dfrac{2}{{13}}} \right)} \right].x\)

\(M = \left( {99 + 1} \right).x = 100x\)

Thay \(x =  - 8\dfrac{7}{{10}}\) vào \(M\) ta được:

\(M = 100.\left( { - 8\dfrac{7}{{10}}} \right)\) \( = 100.\left( { - \dfrac{{87}}{{10}}} \right) =  - 870\)

\(\begin{array}{l}6\dfrac{1}{3}:4\dfrac{2}{9} < x < \left( {10\dfrac{2}{9} + 2\dfrac{2}{5}} \right) - 6\dfrac{2}{9}\\\dfrac{{19}}{3}:\dfrac{{38}}{9} < x < \dfrac{{92}}{9} + \dfrac{{12}}{5} - \dfrac{{56}}{9}\\\dfrac{3}{2} < x < \dfrac{{32}}{5}\end{array}\)

Ta có:

\(\begin{array}{l}\dfrac{3}{2} < x < \dfrac{{32}}{5}\\1,5 < x < 6,4\end{array}\)

Vì x là số tự nhiên nên \(x \in \left\{ {2;3;4;5;6} \right\}\).

Hình a: \(2\dfrac{1}{3}\)

Hình b: \(4\dfrac{5}{6}\)

Hình c: \(6\dfrac{1}{6}\)

Hình d: \(9\dfrac{1}{2}\)

Vậy ta được các hỗn số: \(2\dfrac{1}{3}\); \(4\dfrac{5}{6}\); \(6\dfrac{1}{6}\); \(9\dfrac{1}{2}\).

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.

2 giờ 15 phút = \(2 + \dfrac{{15}}{{60}} = 2 + \dfrac{1}{4} = 2\dfrac{1}{4}\) giờ.