Bộ 10 đề thi cuối kì 2 Toán 6 Chân trời sáng tạo có đáp án - Đề 10

Tìm x, biết: a) x + 1,05 = 4,25 - 0,2   b) (x - 7)/3 =- 1/5

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Tìm \(x\), biết:

    a) \[x + 1,05 = 4,25 - 0,2\];     b) \(\frac{{x - 7}}{3} = \frac{{ - 1}}{5}\);     c) \(\frac{5}{6} + \left( {5x + \frac{3}{2}} \right):\frac{8}{{15}} = 2\frac{1}{{12}}\).

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Giải thích

a) \[x + 1,05 = 4,25 - 0,2\]

\[x + 1,05 = 4,05\]

\[x = 4,05 - 1,05\]

\[x = 3\].

Vậy \[x = 3\].

b) \(\frac{{x - 7}}{3} = \frac{{ - 1}}{5}\)

\(\frac{{5\left( {x - 7} \right)}}{{15}} = \frac{{ - 3}}{{15}}\)

\(5\left( {x - 7} \right) =  - 3\)

\(5x - 35 =  - 3\)

\(5x =  - 3 + 35\)

\(5x = 32\)

\(x = \frac{{32}}{5}\)

Vậy \(x = \frac{{32}}{5}\).

c) \(\frac{5}{6} + \left( {5x + \frac{3}{2}} \right):\frac{8}{{15}} = 2\frac{1}{{12}}\)

\(\left( {5x + \frac{3}{2}} \right):\frac{8}{{15}} = \frac{{25}}{{12}} - \frac{5}{6}\)

\(\left( {5x + \frac{3}{2}} \right):\frac{8}{{15}} = \frac{{15}}{{12}}\)

\(5x + \frac{3}{2} = \frac{5}{4}.\frac{8}{{15}}\)

\(5x + \frac{3}{2} = \frac{2}{3}\)

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

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

  \(x =  - \frac{1}{6}\)

Vậy \(x =  - \frac{1}{6}\).