I. PHẦN TRẮC NGHIỆM KHÁCH QUAN (3,0 điểm)

Hãy khoanh tròn vào phương án đúng duy nhất trong mỗi câu dưới đây.

Cho các số hữu tỉ: \(1,3; - 2,5;\frac{5}{7}; - 1;\frac{{ - 3}}{4}\). Trong các số đã cho có bao nhiêu số hữu tỉ âm?

Ta có \(A = \frac{{4n - 1}}{{n + 2}} = \frac{{4n + 8 - 9}}{{n + 2}} = \frac{{4\left( {n + 2} \right) - 9}}{{n + 2}} = 4 - \frac{9}{{n + 2}}\).

Để biểu thức \(A\) là số nguyên thì \(\frac{9}{{n + 2}}\) là số nguyên hay \(9\,\, \vdots \,\,\left( {n + 2} \right)\).

Suy ra \(\left( {n + 2} \right) \in \)Ư(9) Hay Ư(9) \( = \left\{ { \pm 1; \pm 3; \pm 9} \right\}\).

Ta có bảng sau:

Vậy để biểu thức\(A\) nhận giá trị nguyên thì \(n \in \left\{ { - 11;\,\, - 5;\,\, - 3;\,\, - 1;\,\,1;\,\,7} \right\}\) .

1.

a) \(\left( {\frac{3}{5} - \frac{3}{4}x} \right):\frac{7}{5} = - \frac{1}{2}\)

\(\frac{3}{5} - \frac{3}{4}x = - \frac{1}{2}.\frac{7}{5}\)

\(\frac{3}{5} - \frac{3}{4}x = \frac{{ - 7}}{{10}}\)

\(\frac{3}{4}x = \frac{3}{5} - \frac{{ - 7}}{{10}}\)

\(\frac{3}{4}x = \frac{{13}}{{10}}\)

\[x = \frac{{13}}{{10}}:\frac{3}{4}\]

\[x = \frac{{26}}{{15}}\]

Vậy \[x = \frac{{26}}{{15}}\].

b) \(\left| {\frac{3}{2}x - \frac{1}{6}} \right| - {\left( {\frac{{ - 3}}{2}} \right)^2} = \frac{1}{3}\)

\(\left| {\frac{3}{2}x - \frac{1}{6}} \right| - \frac{9}{4} = \frac{1}{3}\)

\(\left| {\frac{3}{2}x - \frac{1}{6}} \right| = \frac{1}{3} + \frac{9}{4}\)

\(\left| {\frac{3}{2}x - \frac{1}{6}} \right| = \frac{{31}}{{12}}\)

Trường hợp 1: \(\frac{3}{2}x - \frac{1}{6} = \frac{{31}}{{12}}\)

\(\frac{3}{2}x = \frac{{31}}{{12}} + \frac{1}{6}\)

\(\frac{3}{2}x = \frac{{11}}{4}\)

\(x = \frac{{11}}{4}:\frac{3}{2}\)

\(x = \frac{{11}}{6}\)

Trường hợp 2: \(\frac{3}{2}x - \frac{1}{6} = \frac{{ - 31}}{{12}}\)

\(\frac{3}{2}x = \frac{{ - 31}}{{12}} + \frac{1}{6}\)

\(\frac{3}{2}x = \frac{{ - 29}}{{12}}\)

\(x = \frac{{ - 29}}{{12}}:\frac{3}{2}\)

\(x = \frac{{ - 29}}{{18}}\)

Vậy \(x \in \left\{ {\frac{{11}}{6};\,\,\frac{{ - 29}}{{18}}} \right\}\).

2.

a) \(A = \frac{7}{{38}}\,\,.\,\,\frac{9}{{11}} + \frac{7}{{38}}\,\,.\,\,\frac{4}{{11}} - \left| {\frac{{ - 7}}{{38}}} \right|\,\,.\,\,\frac{2}{{11}}\)\( = \frac{7}{{38}}\,\,.\,\,\frac{9}{{11}} + \frac{7}{{38}}\,\,.\,\,\frac{4}{{11}} - \frac{7}{{38}}\,\,.\,\,\frac{2}{{11}}\)

\( = \frac{7}{{38}}\,\,.\,\,\left( {\frac{9}{{11}} + \frac{4}{{11}} - \frac{2}{{11}}} \right)\)\( = \frac{7}{{38}}\,\,.\,\,\frac{{11}}{{11}}\)\( = \frac{7}{{38}}\).

b) \(B = \sqrt {\frac{{81}}{{25}}} \,\,.\,\,{\left( {\frac{{ - 5}}{3}} \right)^3} - \left| {\frac{{ - 12}}{7}} \right|:{\left( {\frac{{ - 3}}{7}} \right)^2} - \frac{{12}}{3}\)\( = \frac{9}{5}.\frac{{ - 125}}{{27}} - \frac{{12}}{7}:\frac{9}{{49}} - \frac{{12}}{3}\)

\( = \frac{9}{5}.\frac{{5.\left( { - 25} \right)}}{{9.3}} - \frac{{12}}{7}.\frac{{49}}{9} - \frac{{12}}{3}\)\( = \frac{{9.5.\left( { - 25} \right)}}{{5.9.3}} - \frac{{3.4.7.7}}{{7.3.3}} - \frac{{12}}{3}\)

\( = \frac{{ - 25}}{3} - \frac{{4.7}}{3} - \frac{{12}}{3}\)\( = \frac{{ - 65}}{3}\).