Giải bài 1 trang 19 vở thực hành Toán 7

Đề bài

Tìm \(x\), biết:

a,\(2x + \frac{1}{2} = \frac{7}{9};\)                                     b,\(\frac{3}{4} - 6x = \frac{7}{{13}}.\)

Lời giải chi tiết

a, \(\begin{array}{l}2x + \frac{1}{2} = \frac{7}{9}\\ \Leftrightarrow 2x = \frac{7}{9} - \frac{1}{2}\\ \Leftrightarrow 2x = \frac{{14}}{{18}} - \frac{9}{{18}}\\ \Leftrightarrow 2x = \frac{5}{{18}}\\ \Leftrightarrow x = \frac{5}{{18}}:2\\ \Leftrightarrow x = \frac{5}{{18}}.\frac{1}{2}\\ \Leftrightarrow x = \frac{5}{{36}}\end{array}\)

b, \(\begin{array}{l}\frac{3}{4} - 6x = \frac{7}{{13}}\\ \Leftrightarrow 6x = \frac{3}{4} - \frac{7}{{13}}\\ \Leftrightarrow 6x = \frac{{39}}{{52}} - \frac{{28}}{{52}}\\ \Leftrightarrow 6x = \frac{{11}}{{52}}\\ \Leftrightarrow x = \frac{{11}}{{52}}:6\\ \Leftrightarrow x = \frac{{11}}{{52}}.\frac{1}{6}\\ \Leftrightarrow x = \frac{{11}}{{312}}\end{array}\)