Tìm x , biết: a) 3/7 + 4/ 7 x = 1/ 3 ; b) − 4 /x = x /− 9 ;
a) \(\frac{3}{7} + \frac{4}{7}x = \frac{1}{3}\)
\(\frac{4}{7}x = \frac{1}{3} - \frac{3}{7}\)
\(\frac{4}{7}x = \frac{{ - 2}}{{21}}\)
\(x = \frac{{ - 2}}{{21}}:\frac{4}{7}\)
\(x = \frac{{ - 1}}{6}\)
Vậy \(x = - \frac{1}{6}\).
b) \(\frac{{ - 4}}{x} = \frac{x}{{ - 9}}\)
\({x^2} = \left( { - 4} \right).\left( { - 9} \right)\)
\({x^2} = 36\)
Suy ra \(x = 6\) hoặc \(x = - 6\)
Vậy \(x \in \left\{ {6; - 6} \right\}\).
c) \(\frac{5}{{11}} + \frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{4}{5}\)
\(\frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{4}{5} - \frac{5}{{11}}\)
\(\frac{2}{{11}}:\left| {\frac{3}{7} - x} \right| = \frac{{19}}{{55}}\)
\(\left| {\frac{3}{7} - x} \right| = \frac{2}{{11}}:\frac{{19}}{{55}}\)
\(\left| {\frac{3}{7} - x} \right| = \frac{{10}}{{19}}\)
Trường hợp 1: \(\frac{3}{7} - x = \frac{{10}}{{19}}\) \(x = \frac{3}{7} - \frac{{10}}{{19}}\) \(x = - \frac{{13}}{{133}}\) Vậy \(x \in \left\{ { - \frac{{13}}{{133}};\,\frac{{127}}{{133}}} \right\}\). | Trường hợp 2: \(\frac{3}{7} - x = - \frac{{10}}{{19}}\) \(x = \frac{3}{7} + \frac{{10}}{{19}}\) \(x = \frac{{127}}{{133}}\) |