Bộ 10 đề thi giữa kì 2 Toán 6 Cánh diều có đáp án - Đề 06

1. Thực hiện phép tính (tính nhanh nếu có thể): a) (- 1/6 + 5/ - 12) + 7/12; 2. Tìm \[x\]: a) 11/8- 3/8.x =1/8;

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II. PHẦN TỰ LUẬN

1. Thực hiện phép tính (tính nhanh nếu có thể):

a) \[\left( {\frac{{ - 1}}{6} + \frac{5}{{ - 12}}} \right) + \frac{7}{{12}}\];                                    b) \[\frac{7}{{36}} - \frac{8}{{ - 9}} + \frac{{ - 2}}{3}\];

2. Tìm \[x\]:

a) \[\frac{{11}}{8} - \frac{3}{8} \cdot x = \frac{1}{8}\];                                          b) \({\left( {{\rm{x}} - \frac{1}{2}} \right)^{\rm{2}}}{\rm{ = }}\frac{{\rm{1}}}{4}\).

0/3000 ký tự
Giải thích

1.

a) \[\left( {\frac{{ - 1}}{6} + \frac{5}{{ - 12}}} \right) + \frac{7}{{12}}\]\[ = \frac{{ - 1}}{6} + \left( {\frac{5}{{ - 12}} + \frac{7}{{12}}} \right)\]

\[ = \frac{{ - 1}}{6} + \left( {\frac{{ - 5}}{{12}} + \frac{7}{{12}}} \right)\]\[ = \frac{{ - 1}}{6} + \frac{2}{{12}}\]\[ = \frac{{ - 1}}{6} + \frac{1}{6} = 0\];

b) \[\frac{7}{{36}} - \frac{8}{{ - 9}} + \frac{{ - 2}}{3}\]\[ = \frac{7}{{36}} + \frac{8}{9} + \frac{{ - 2}}{3}\]

\[ = \frac{7}{{36}} + \frac{{32}}{{36}} + \frac{{ - 24}}{{36}}\]\( = \frac{{15}}{{36}} = \frac{5}{{12}}\).

2.  

a) \[\frac{{11}}{8} - \frac{3}{8} \cdot x = \frac{1}{8}\]

\[\frac{3}{8} \cdot x = \frac{{11}}{8} - \frac{1}{8}\]

\[\frac{3}{8} \cdot x = \frac{{10}}{8}\]

\[x = \frac{{10}}{8}:\frac{3}{8}\]

\[x = \frac{{10}}{3}\].

Vậy \[x = \frac{{10}}{3}\].

b) \({\left( {x - \frac{1}{2}} \right)^2} = \frac{1}{4}\)

• TH1: \(x - \frac{1}{2} = \frac{1}{2}\)

\(x = \frac{1}{2} + \frac{1}{2}\)

\(x = 1\).

 

• TH2: \(x - \frac{1}{2} = \frac{{ - 1}}{2}\)\(x = \frac{{ - 1}}{2} + \frac{1}{2}\)

\(x = 0\).

Vậy \(x = 1\,;\,\,\,x = 0\).