Đề thi Giữa kì 2 Toán 6 trường THCS&THPT Lương Thế Vinh (Hà Nội) năm 2023-2024 có đáp án

Đáp án đúng là: C

Ta có \(\frac{{ - 18}}{{24}} = \frac{{ - 3}}{4} = \frac{{12}}{{ - 16}} = - \frac{{15}}{{20}}.\)

Đáp án đúng là: C

Từ \(\frac{x}{4} = \frac{{ - 50}}{{ - 25}}\)

Suy ra \(x = \frac{{4.\left( { - 50} \right)}}{{ - 25}} = 8\).

Đáp án đúng là: B

Sau \(1,5\) giờ anh đi được số km là: \(\frac{2}{9}.45.1,5 = 15\left( {km} \right)\).

Đáp án đúng là: D

Ta có \(\frac{{AI}}{{BI}} = \frac{3}{5}\) suy ra \(\frac{{BI}}{{AB}} = \frac{5}{8}\)

Do đó \(BI = \frac{5}{8}AB = \frac{5}{8}.16 = 10\left( {cm} \right)\).

a) \[\left( { - 6,4} \right) + \left( { - 8,3} \right) + 6,4 - 1,7\]

\[ = \left[ {\left( { - 6,4} \right) + 6,4} \right] + \left( { - 8,3 - 1,7} \right)\]

\[ = 0 + \left( { - 10} \right)\]\[ = - 10\].

b) \(\frac{7}{{15}}.1\frac{{14}}{{33}} + \frac{{19}}{{33}}:\frac{{15}}{7} - \frac{7}{{15}}\)

\( = \frac{7}{{15}}.1\frac{{14}}{{33}} + \frac{{19}}{{33}}.\frac{7}{{15}} - \frac{7}{{15}}.1\)

\( = \frac{7}{{15}}.\left( {1\frac{{14}}{{33}} + \frac{{19}}{{33}} - 1} \right)\)

\( = \frac{7}{{15}}.1\)\( = \frac{7}{{15}}\).

c) \(\frac{1}{2} + \frac{{35}}{{78}} - 1\frac{5}{{37}} - \left( {1\frac{{35}}{{78}} + \frac{{32}}{{37}} - 2,5} \right)\)

\( = \frac{1}{2} + \frac{{35}}{{78}} - 1\frac{5}{{37}} - 1\frac{{35}}{{78}} - \frac{{32}}{{37}} + 2,5\)

\( = \left( {\frac{1}{2} + \frac{5}{2}} \right) + \left( {\frac{{35}}{{78}} - 1\frac{{35}}{{78}}} \right) - \left( {1\frac{5}{{37}} + \frac{{32}}{{37}}} \right)\)

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

\( = 0\).

a) \(\frac{4}{5} - x = \frac{6}{{15}}\)

\(x = \frac{4}{5} - \frac{6}{{15}}\)

\(x = \frac{2}{5}\)

Vậy \(x = \frac{2}{5}\).

b) \(0,6x + \frac{4}{5}\left( {x - 10} \right) = - 7\)

\(\frac{3}{5}x + \frac{4}{5}x - 8 = - 7\)

\(\left( {\frac{3}{5} + \frac{4}{5}} \right)x = - 7 + 8\)

\(\frac{7}{5}x = 1\)

\(x = 1:\frac{7}{5}\)

\(x = \frac{5}{7}\)

Vậy \(x = \frac{5}{7}\).

c) \(0,5 - \frac{3}{4}{x^2} = \frac{{ - 5}}{2}\)

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

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

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

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

\({x^2} = 4\)

\(x = \pm 2\)

Vậy \(x \in \left\{ { \pm 2} \right\}\).