Question

Решите 1 номер пожалуйста

Answer 1

$1) f(x)=sqrt[4]{3-2x-x^2},$

ООФ: $3-2x-x^2ge0,$

$-x^2-2x+3=0,\D=(-2)^2-4bullet(-1)bullet3=4+12=16,\x_{1,2}=frac{-(-2)pmsqrt{16}}{-2}=frac{2pm4}{-2}=-1mp2,\x_1=-1-2=-3,\x_2=-1+2=1,\-x^2-2x+3=-(x+3)(x-1)ge0,\a=-1 textless 0,to -3le xle1.\\OTBET: -3le xle 1.$

$2) f(x)=sqrt{3-2x}-sqrt[6]{9-x^2},$

ООФ: $left { {{3-2xge0} atop {9-x^2ge0}} right,$

$a) 3-2xge0,\2xle3,\xlefrac{3}{2},\xle1.5;\\b) 9-x^2ge0,\(3-x)(3+x)ge0,\a=-1to -3le xle3.\\ left { {{xle1,5,} atop -3le xle3} right. to -3le xle1,5.\\OTBET: -3le xle1,5.$