Question

Вычислить(Еще необходимо написать ОДЗ)

1)$sqrt{ x^{2} + 45 } = sqrt{18x}$

2)$sqrt{3 x^{2} -12x + 12} = x - 2$

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

Answer 1

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