Tìm \(x\), biết:
a) \(\frac{6}{7} - x = \frac{{12}}{{28}}.\)
b) \(\frac{{ - 5}}{3} + \frac{7}{{10}}x = 0,2.\)
c) \({\left( {2x + 6} \right)^2} = \frac{{81}}{{25}}.\)
a), b) Chuyển vế để tìm x.
c) Với \({A^2} = {B^2}\), ta chia hai trường hợp: TH1: A = B; TH2: A = - B.
a) \(\frac{6}{7} - x = \frac{{12}}{{28}}\)
\(x = \frac{6}{7} - \frac{{12}}{{28}}\)
\(x = \frac{{24}}{{28}} - \frac{{12}}{{28}}\)
\(x = \frac{{12}}{{28}} = \frac{3}{7}\)
Vậy \(x = \frac{3}{7}\)
b) \(\frac{{ - 5}}{3} + \frac{7}{{10}}x = 0,2\)
\(\frac{7}{{10}}x = \frac{1}{5} + \frac{5}{3}\)
\(\frac{7}{{10}}x = \frac{3}{{15}} + \frac{{25}}{{15}} = \frac{{28}}{{15}}\)
\(x = \frac{{28}}{{15}}:\frac{7}{{10}}\)
\(x = \frac{8}{3}\)
Vậy \(x = \frac{8}{3}\)
c) \({\left( {2x + 6} \right)^2} = \frac{{81}}{{25}}\)
\({\left( {2x + 6} \right)^2} = {\left( {\frac{9}{5}} \right)^2} = {\left( {\frac{{ - 9}}{5}} \right)^2}.\)
* TH1: \(2x + 6 = \frac{9}{5}\)
\(2x = \frac{9}{5} - 6\)
\(2x = \frac{{ - 21}}{5}\)
\(x = \frac{{ - 21}}{5}:2\)
\(x = \frac{{ - 21}}{{10}}.\)
* TH2: \(2x + 6 = \frac{{ - 9}}{5}\)
\(2x = \frac{{ - 9}}{5} - 6\)
\(2x = \frac{{ - 39}}{5}\)
\(x = \frac{{ - 39}}{5}:2\)
\(x = \frac{{ - 39}}{{10}}.\)
Vậy \(x \in \left\{ {\frac{{ - 39}}{{10}};\frac{{ - 21}}{{10}}} \right\}\).