Question

Решите четные задания.

Answer 1

$8\2);frac{40}{x-18}-frac{40}{x+2}=1\frac{40(x+2)-40(x-18)}{(x-18)(x+2)}=1\frac{40x+80-40x+720}{x^2-16x-36}=1\x^2-16x-36=800\x^2-16x-836=0\D=256+4cdot836=3600=(60)^2\x_{1,2}=frac{16pm60}2\x_1=22,;x_2=38\\4);frac1{3-x}=frac3{20}+frac1x\frac1{3-x}=frac{3x+20}{20x}\20x=(3x+20)(3-x)\20x=-3x^2-11x+60\-3x^2-31x+60=0\3x^2+31x-60=0\D=961+4cdot3cdot60=1681=(41)^2\x_{1,2}=frac{-31pm41}6\x_1=-12,;x_2=frac{10}6=1frac23$

$6);frac2{x-1}+frac2{x+1}=frac32\frac{2x+2+2x-2}{x^2-1}=frac32\8x=3x^2-3\3x^2-8x-3=0\D=64+4cdot3cdot3=100=(10)^2\x_{1,2}=frac{8pm10}6\x_1=-frac13,;x_2=3\\11.\2);9x^4-37x^2+4=0\x^2=t,;x^4=t^2,;tgeq0\9t^2-37t+4=0\D=1369-4cdot9cdot4=1225=(35)^2\t_{1,2}=frac{37pm35}{18}\t_1=frac19,;t_2=4\x=sqrt t\x_1=-frac13,;x_2=frac13,;x_3=-2,;x_4=2$

$4);(x+3)^4+(x+3)^2-20=0\(x+3)^2=t,;(x+3)^4=t^2,;tgeq0\t^2+t-20=0\D=1+4cdot20=81=(9)^2\t_{1,2}=frac{-1pm9}2\t_1=-5;-;He;nogx.\t_2=4\x=sqrt t\x_2=-2,;x_2=2$