Question

решите уравнение....​

Answer 1

$\displaystyle\bf\\\frac{1}{x} +\frac{1}{x-3} =\frac{1}{2} \\\\x\neq 0 \ \ ; \ \ x\neq 3\\\\\\\frac{x-3+x}{x\cdot(x-3)} =\frac{1}{2} \\\\\\\frac{2x-3}{x\cdot(x-3)} =\frac{1}{2} \\\\\\x\cdot(x-3)=2\cdot(2x-3)\\\\x^{2} -3x=4x-6\\\\x^{2} -3x-4x+6=0\\\\x^{2} -7x+6=0\\\\D=(-7)^{2} -4\cdot 6=49-24=25=5^{2} \\\\\\x_{1} =\frac{7-5}{2} =1\\\\\\x_{2} =\frac{7+5}{2} =6\\\\\\Otvet \ : \ 1 \ ; \ 6$