Đề bài

Tìm \(x\):

a) \(x - \frac{1}{4} = \frac{2}{3}\)

b) \(\frac{1}{4} + \frac{3}{4}x = \frac{{ - 13}}{8}\)

c) \(\left| {\frac{3}{4}x - \frac{1}{2}} \right| + {\left( {\frac{{ - 1}}{2}} \right)^2} = \sqrt {\frac{4}{9}} \)

Phương pháp giải

Áp dụng quy tắc chuyển vế đổi dấu.

c) Đưa về dạng \(\left| A \right| = B\), chia hai trường hợp: A = B hoặc A = -B.

Lời giải của GV Loigiaihay.com

a) \(x - \frac{1}{4} = \frac{2}{3}\)

\(\begin{array}{l}x = \frac{2}{3} + \frac{1}{4}\\x = \frac{{11}}{{12}}\end{array}\)

Vậy \(x = \frac{{11}}{{12}}\)

b) \(\frac{1}{4} + \frac{3}{4}x = \frac{{ - 13}}{8}\)

\(\begin{array}{l}\frac{3}{4}x = \frac{{ - 13}}{8} - \frac{1}{4}\\\frac{3}{4}x = \frac{{ - 15}}{8}\\x = \frac{{ - 15}}{8}:\frac{3}{4}\\x = \frac{{ - 5}}{2}\end{array}\)

Vậy \(x = \frac{{ - 5}}{2}\)

c) \(\left| {\frac{3}{4}x - \frac{1}{2}} \right| + {\left( {\frac{{ - 1}}{2}} \right)^2} = \sqrt {\frac{4}{9}} \)

\(\begin{array}{l}\left| {\frac{3}{4}x - \frac{1}{2}} \right| + \frac{1}{4} = \frac{2}{3}\\\left| {\frac{3}{4}x - \frac{1}{2}} \right| = \frac{2}{3} - \frac{1}{4}\\\left| {\frac{3}{4}x - \frac{1}{2}} \right| = \frac{5}{{12}}\\\frac{3}{4}x - \frac{1}{2} =  \pm \frac{5}{{12}}\end{array}\)

TH1: \(\frac{3}{4}x - \frac{1}{2} = \frac{5}{{12}}\)

\(\begin{array}{l}\frac{3}{4}x = \frac{5}{{12}} + \frac{1}{2}\\\frac{3}{4}x = \frac{{11}}{{12}}\\x = \frac{{11}}{{12}}:\frac{3}{4}\\x = \frac{{11}}{9}\end{array}\)

TH2: \(\frac{3}{4}x - \frac{1}{2} =  - \frac{5}{{12}}\)

\(\begin{array}{l}\frac{3}{4}x =  - \frac{5}{{12}} + \frac{1}{2}\\\frac{3}{4}x = \frac{1}{{12}}\\x = \frac{1}{{12}}:\frac{3}{4}\\x = \frac{1}{9}\end{array}\)

Vậy \(x \in \left\{ {\frac{{11}}{9};\frac{1}{9}} \right\}\)