Question

ДАЮ 60 БАЛЛОВ.ПОМОГИТЕ СРОЧНО РЕШИТЬ 2,4 С ОБЪЯСНЕНИЕМ

Answer 1

Ответ:

Дробно-рациональные уравнения .

$\displaystyle \bf 2)\ \ \frac{2x-3}{x+1}+\frac{x+1}{2x-3}=2\ \ ,\ \ \ ODZ:\ x\ne -1\ ,\ x\ne 1,5\\\\\\\frac{(2x-3)^2+(x+1)^2}{(x+1)(2x-3)}-2=0\ \ ,\ \ \ \frac{4x^2-12x+9+x^2+2x+1}{(x+1)(2x-3)}-2=0\\\\\\\frac{5x^2-10x+10-2(x+1)(2x-3)}{(x+1)(2x-3)}=0\ \ ,\\\\\\\frac{5x^2-10x+10-4x^2+2x+6}{(x+1)(2x-3)}=0\ \ ,\ \ \ \frac{x^2-8x+16}{(x+1)(2x-3)}=0\ \ ,\\\\\\\frac{(x-4)^2}{(x+1)(2x-3)}=0\ \ \ \Rightarrow \ \ \ (x-4)^2=0\ \ ,\ \ x=4\\\\\\Otvet:\ x=4\ .$  

$\displaystyle \bf 4)\ \ \frac{x^2}{(3x+1)^2}-\frac{6x}{3x+1}+5=0 \ \ ,\ \ \ \ ODZ:\ x\ne -\frac{1}{3}\ \ ,\\\\\\\frac{x^2-6x(3x+1)+5(3x+1)^2}{(3x+1)^2}=0\ \ ,\\\\\\\frac{x^2-18x^2-6x+5(9x^2+6x+1)}{(3x+1)^2}=0\ \ ,\\\\\\\frac{-17x^2-6x+45x^2+30x+5}{(3x+1)^2}=0\ \ ,\\\\\\\frac{28x^2+24x+5}{(3x+1)^2}=0\ \ \ \Rightarrow \ \ \ 28x^2+24x+5=0\ \ ,\\\\\\D/4=(b/2)^2-ac=12^2-28\cdot 5=4\ \ ,\\\\\\x_1=\frac{-12-2}{28}=-\frac{1}{2}\ \ ,\ \ \ x_1=\frac{-12+2}{28}=-\frac{5}{14}$  

$\bf Otvet:\ x_1=-\dfrac{1}{2}\ ,\ x_2=-\dfrac{5}{14}\ .$