Bộ 10 đề thi cuối kì 1 Toán 7 Chân trời sáng tạo có đáp án - Đề 5
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Tìm x của các câu sau , biết: a) ( 3 /5 − 3/ 4 x ) : 7/ 5 = − 1/ 2 ; b) ∣ ∣ 3/2 x − 1/6 ∣ ∣ − ( − 3 /2 )^2 = 1/3

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II. PHẦN TỰ LUẬN (7,0 điểm)

 (2,0 điểm)

1. Tìm \(x\), biết:

a) \(\left( {\frac{3}{5} - \frac{3}{4}x} \right):\frac{7}{5} = - \frac{1}{2}\);                                                            b) \(\left| {\frac{3}{2}x - \frac{1}{6}} \right| - {\left( {\frac{{ - 3}}{2}} \right)^2} = \frac{1}{3}\)

2. Thực hiện phép tính (tính hợp lí nếu có thể):

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

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

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Giải thích

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

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

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