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1) $sqrt{ x^{2} + 45 } = sqrt{18x}$
2) $sqrt{3 x^{2} -12x + 12} = x - 2$
3) $(2 - sqrt{2x - 4} )( sqrt{3x - 11 - 2} ) = 0$

1)
ОДЗ:
$x^{2} +45 geq 0 x^{2} geq -45 x in R 18x geq 0 x geq frac{0}{18} x geq 0$

$(sqrt{ x^{2} + 45 })^2 = (sqrt{18x} )^2 x^{2} +45=18x x^{2} -18x+45=0 D=(-18)^2-4*1*45=324-180=144=12^2 x_1= frac{-(-18)- sqrt{144} }{2*1} = frac{18-12}{2} = frac{6}{2} =3 x_2= frac{-(-18)+ sqrt{144} }{2*1} = frac{18+12}{2} = frac{30}{2} =15$

Ответ: х=15 и х=3

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2)
ОДЗ:
$3 x^{2} -12x+12 geq 0 3 x^{2} -12x+12 =0 D=(-12)^2-4*12*3=144-144=0 x= frac{-(-12)}{2*3} = frac{12}{6} =2 x geq 2 x-2 geq 0 x geq 2$

$sqrt{3 x^{2} -12x + 12} = x - 2 (sqrt{3 x^{2} -12x + 12})^2 =( x - 2)^2 mathrm{(a-b)^2=a^2-2ab+b^2} 3x^2-12x+12= x^{2} -4x+4 3 x^{2} - x^{2} -12x+4x+12-4=0 2 x^{2} -8x+8=0 2( x^{2} -4x+4)=0 x^{2} -4x+4=0 D=4^2-4*4*1=16-16=0 x= frac{-(-4)}{2*1} = frac{4}{2} =2$

Ответ: х=2

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3)
ОДЗ:
$2x-4 geq 0 2x geq 4 x geq frac{4}{2} x geq 2 3x-11-2 geq 0 3x-13 geq 0 3x geq 13 x geq frac{13}{3} x geq 4 frac{1}{3}$

$(2 - sqrt{2x - 4} )( sqrt{3x - 11 - 2} ) = 0 2 - sqrt{2x - 4}=0 sqrt{2x - 4}=2 (sqrt{2x - 4})^2=2^2 2x-4=4 2x=4-4 2x=0 x=0 textless 2$

$sqrt{3x - 11 - 2} = 0 (sqrt{3x - 11 - 2})^2 = 0^2 3x-11-2=0 3x-13=0 3x =13 x =frac{13}{3} x =4 frac{1}{3}$

Ответ: x= 4 1/3