Question

Решите уровнение под номером 9.
По фото

Answer 1

$\displaystyle \tt \frac{1}{(x-2)^2}-\frac{1}{x-2}-6=0\:\:\:\:\: | \: x\ne2\\\\ \displaystyle \tt \frac{1-(x-2)-6(x-2)^2}{(x-2)^2}=0\\\\ \displaystyle \tt 1-x+2-6(x^2-4x+4)=0\\\displaystyle \tt 3-x-6x^2+24x-24=0\\\displaystyle \tt -6x^2+23x-21=0\:\:\:\:\:|\cdot (-1)\\\displaystyle \tt 6x^2-23x+21=0\\\displaystyle \tt D=(-23)^2-4\cdot6\cdot21=529-504=25\\\displaystyle \tt \sqrt{D}=\sqrt{25}=5$

$\displaystyle \tt \bold{x_1}=\frac{23+5}{2\cdot6}=\frac{28}{12}=\frac{7}{3}=\bold{2\frac{1}{3}}\\\\ \displaystyle \tt \bold{x_2}=\frac{23-5}{2\cdot6}=\frac{18}{12}=\frac{3}{2}=\bold{1,5}$

Answer 2

$\dfrac{1}{(x-2)^2}-\dfrac{1}{x-2}-6=0\\ t=\dfrac{1}{x-2}\\ t^2-t-6=0\\ t_1+t_2=1,t_1t_2=-6=>t=3\;\;\;\;\;\;t=-2\\ \dfrac{1}{x-2}=3=>x-2=\dfrac{1}{3}=>x=2\dfrac{1}{3}\\ \dfrac{1}{x-2}=-2=>x-2=-\dfrac{1}{2}=>x=1\dfrac{1}{2}\\ OTBET: 2\dfrac{1}{3},1\dfrac{1}{2}$