Решите уравнение)))) 10/x+4/(x-3)=1

[latex] \frac{10}{x} + \frac{4}{x-3} =1 \\ \\ 10(x-3)+4x=x(x-3) \\ 10x-30+4x= x^{2} -3x \\ 14x-30= x^{2} -3x \\ x^{2} -3x-14x+30=0 \\ x^{2} -17x+30=0 \\ \\ D=289-120=169 \\ \sqrt{D} =13 \\ x1= \frac{17+13}{2} =15 \\ x2= \frac{17-13}{2} =2 \\ \\ x \neq 0 \\ x \neq 3[/latex] Ответ:х=15;х=2.

[latex]\frac{10}{x}+\frac{4}{x-3}=1;\ x\neq0\ \wedge\ x-3\neq0\to x\neq0\ \wedge\ x\neq3\\\\\frac{10(x-3)}{x(x-3)}+\frac{4x}{x(x-3)}=1\\\\\frac{10x-30+4x}{x^2-3x}=1\\\\\frac{14x-30}{x^2-3x}=1\iff14x-30=x^2-3x\\\\x^2-3x-14x+30=0\\\\x^2-17x+30=0[/latex] [latex]x^2-2x-15x+30=0\\\\x(x-2)-15(x-2)=0\\\\(x-2)(x-15)=0\iff x-2=0\ \vee\ x-15=0\\\\x=2\ \vee\ x=15[/latex]