Giải bài 5 trang 30 SGK Toán 7 tập 1 – Cánh diều

Đề bài

Tìm x, biết:

a) \(x + \left( { – \frac{2}{9}} \right) = \frac{{ – 7}}{{12}}\);

b) \(( – 0,1) – x = \frac{{ – 7}}{6}\)

c) \(( – 0,12) \cdot \left( {x – \frac{9}{{10}}} \right) =  – 1,2\);

d) \(\left( {x – \frac{3}{5}} \right):\frac{{ – 1}}{3} = 0,4.\)

Lời giải chi tiết

a)

\(\begin{array}{l}x + \left( { – \frac{2}{9}} \right) = \frac{{ – 7}}{{12}}\\x = \frac{{ – 7}}{{12}} + \frac{2}{9}\\x = \frac{{ – 21}}{{36}} + \frac{8}{{36}}\\x = \frac{{ – 13}}{{36}}\end{array}\)

Vậy \(x = \frac{{ – 13}}{{36}}\).

b)

\(\begin{array}{l}( – 0,1) – x = \frac{{ – 7}}{6}\\\frac{{ – 1}}{{10}} – x = \frac{{ – 7}}{6}\\x = \frac{{ – 1}}{{10}} + \frac{7}{6}\\x = \frac{{ – 3}}{{30}} + \frac{{35}}{{30}}\\x = \frac{{32}}{{30}}\\x = \frac{{16}}{{15}}\end{array}\)

Vậy \(x = \frac{{16}}{{15}}\)

c)

\(\begin{array}{l}( – 0,12) \cdot \left( {x – \frac{9}{{10}}} \right) =  – 1,2\\\frac{{ – 3}}{{25}} \cdot \left( {x – \frac{9}{{10}}} \right) = \frac{{ – 6}}{5}\\x – \frac{9}{{10}} = \frac{{ – 6}}{5}:\left( {\frac{{ – 3}}{{25}}} \right)\\x – \frac{9}{{10}} = \frac{{ – 6}}{5}.\frac{{ – 25}}{3}\\x – \frac{9}{{10}} = 10\\x = 10 + \frac{9}{{10}}\\x = \frac{{109}}{{10}}\end{array}\)

Vậy \(x = \frac{{109}}{{10}}\).

d)

 \(\begin{array}{l}\left( {x – \frac{3}{5}} \right):\frac{{ – 1}}{3} = 0,4\\\left( {x – \frac{3}{5}} \right):\frac{{ – 1}}{3} = \frac{2}{5}\\x – \frac{3}{5} = \frac{2}{5}.\frac{{ – 1}}{3}\\x – \frac{3}{5} = \frac{{ – 2}}{{15}}\\x = \frac{{ – 2}}{{15}} + \frac{3}{5}\\x= \frac{{ – 2}}{{15}} + \frac{9}{15}\\x = \frac{7}{{15}}\end{array}\)

Vậy \(x = \frac{7}{{15}}\).