Вычислить(Еще необходимо написать ОДЗ) 1)[latex] \sqrt{ x^{2} + 45 } = \sqrt{18x} [/latex] 2)[latex] \sqrt{3 x^{2} -12x + 12} = x - 2[/latex] 3)[latex](2 - \sqrt{2x - 4} )( \sqrt{3x - 11 - 2} ) = 0[/latex]

1) ОДЗ: [latex] x^{2} +45 \geq 0 \\ x^{2} \geq -45 \\ x \in R\\ \\ 18x \geq 0 \\ x \geq \frac{0}{18} \\ x \geq 0 [/latex] [latex] (\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 \\ [/latex] Ответ: х=15 и х=3 ________________________________________________ 2) ОДЗ: [latex]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[/latex] [latex] \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[/latex] Ответ: х=2 _____________________________________________________________ 3) ОДЗ: [latex]\\ 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} [/latex] [latex](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 \\ [/latex] [latex]\\ \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} [/latex] Ответ: x= 4 1/3