Question

СРОЧНО!!!!!даю 30 баллов , второе не надо

Answer 1

$3.\;\frac{2x^2+7x-4}{x^2+6x-2}=1\\\\2x^2+7x-4=x^2+6x-2\\x^2+x-2=0\\D=1-4\cdot1\cdot(-2)=1+8=9\\x_{1,2}=\frac{-1\pm3}2\\x_1=-2,\;x_2=1\\\\x_1+x_2=-2+1=-1$

$4.\;\frac1{x-5}+\frac3{x+1}=\frac{x^2-2x-21}{x^2-4x-5}\\\\\frac{x+1+3x-15}{(x-5)(x+1)}=\frac{x^2-2x-21}{x^2-4x-5}\\\\\frac{4x-14}{x^2-4x-5}=\frac{x^2-2x-21}{x^2-4x-5}\\\\x^2-4x-5\neq0\\(x-5)(x+1)\neq0\\x\neq5,\;x\neq-1\\\\4x-14=x^2-2x-21\\x^2-6x-7=0\\D=36-4\cdot1\cdot(-7)=36+28=64\\x_{1,2}=\frac{6\pm8}2\\x_1=7\\x_2=-1\;-\;He\;nogx.\\OTBET:\;7$

$5.\;\left(\frac{x+3}{x-1}\right)^2+12=64\left(\frac{x-1}{x+3}\right)^2\\O.D.3:\\\begin{cases}x\neq1\\x\neq-3\end{cases}\\\\\left(\frac{x+3}{x-1}\right)^2=t,\;\left(\frac{x-1}{x+3}\right)^2=\frac1t,\;t\geq0\\\\t+12=\frac{64}t\\t^2+12t=64\\t^2+12t-64=0\\D=144+4\cdot1\cdot(-64)=144+256=400\\t_{1,2}=\frac{-12\pm20}2\\t_1=4\\t_2=-16\;-\;He\;nogx.\\\\\left(\frac{x+3}{x-1}\right)^2=4$

$\begin{cases}\frac{x+3}{x-1}=2\\\frac{x+3}{x-1}=-2\end{cases}\Rightarrow\begin{cases}x+3=2x-2\\x+3=2-2x\end{cases}\Rightarrow\begin{cases}x_1=5\\x_2=-\frac13\end{cases}$