Думаю, как-то так:
(2-√3)²- √2(√6-1) +3√12= 2²-2*2*√3-(√3)² -√2*√6+√2 + 3√12=
= 4- 4√3- 3 - √12+ √2+ 3√12= 1 - 4√3 - 2√3 +√2 + 6√3 = 1 +√2
![\frac{ x^{2} -3}{2 \sqrt{3}- 2x } = \frac{(x- \sqrt{3})(x+ \sqrt{3} ) }{2*( \sqrt{3}-x) } = \frac{(x- \sqrt{3})(x+ \sqrt{3} )}{-2(x- \sqrt{3})} =- \frac{x+ \sqrt{3} }{2}](https://tex.z-dn.net/?f=+%5Cfrac%7B+x%5E%7B2%7D+-3%7D%7B2++%5Csqrt%7B3%7D-+2x+%7D+%3D++%5Cfrac%7B%28x-+%5Csqrt%7B3%7D%29%28x%2B+%5Csqrt%7B3%7D+%29+%7D%7B2%2A%28+%5Csqrt%7B3%7D-x%29+%7D+%3D+%5Cfrac%7B%28x-+%5Csqrt%7B3%7D%29%28x%2B+%5Csqrt%7B3%7D+%29%7D%7B-2%28x-+%5Csqrt%7B3%7D%29%7D+%3D-+%5Cfrac%7Bx%2B+%5Csqrt%7B3%7D+%7D%7B2%7D+)
![\frac{y+x-2 \sqrt{x*y} }{5 \sqrt{x} -5 \sqrt{y} } = \frac{( \sqrt{y}- \sqrt{x} )^{2} }{5( \sqrt{x} - \sqrt{y}) } = \frac{( \sqrt{y} - \sqrt{x}) ^{2} }{-5( \sqrt{y}- \sqrt{x} ) } =- \frac{ \sqrt{y}- \sqrt{x} }{5}](https://tex.z-dn.net/?f=+%5Cfrac%7By%2Bx-2+%5Csqrt%7Bx%2Ay%7D+%7D%7B5+%5Csqrt%7Bx%7D+-5+%5Csqrt%7By%7D+%7D+%3D+%5Cfrac%7B%28+%5Csqrt%7By%7D-+%5Csqrt%7Bx%7D+%29%5E%7B2%7D++%7D%7B5%28+%5Csqrt%7Bx%7D+-+%5Csqrt%7By%7D%29+%7D+%3D++%5Cfrac%7B%28+%5Csqrt%7By%7D+-+%5Csqrt%7Bx%7D%29+%5E%7B2%7D++%7D%7B-5%28+%5Csqrt%7By%7D-+%5Csqrt%7Bx%7D+%29+%7D+%3D-+%5Cfrac%7B+%5Csqrt%7By%7D-+%5Csqrt%7Bx%7D++%7D%7B5%7D+)
Смотри, берёшь и подставляешь в уравнение известный корень на место х :
64+8*р+28=0;
8*р=-92;
р=-11,5.
Ответ: -11,5.
И не каких округлений.
-2(5-3x)=7x+3
-(10-6x)=7x+3
-10+6x=7x+3
-10-3=7x-6x
-13=x
x=-13
ОДЗ 2x+8≥0 2x≥-8 x≥-4
![\sqrt{2x+8}=6 \\ (\sqrt{2x+8})^2=6^2 \\ 2x+8=36 \\ 2x=28 \\ x=14](https://tex.z-dn.net/?f=%5Csqrt%7B2x%2B8%7D%3D6+%5C%5C+%28%5Csqrt%7B2x%2B8%7D%29%5E2%3D6%5E2+%5C%5C+2x%2B8%3D36+%5C%5C+2x%3D28+%5C%5C+x%3D14)
ОДЗ x²-4x+13≥0 при любых значениях х
![\sqrt{x^2-4x+13}=5 \\ (\sqrt{x^2-4x+13})^2=5^2 \\ x^2-4x+13=25 \\ x^2-4x-12=0 \\ D=16+48=64 \\ x_1=\frac(4-8}{2}=-2 \ \ \ \ \ x_2=\frac{4+8}{2}=6](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-4x%2B13%7D%3D5+%5C%5C+%28%5Csqrt%7Bx%5E2-4x%2B13%7D%29%5E2%3D5%5E2+%5C%5C+x%5E2-4x%2B13%3D25+%5C%5C+x%5E2-4x-12%3D0+%5C%5C+D%3D16%2B48%3D64+%5C%5C+x_1%3D%5Cfrac%284-8%7D%7B2%7D%3D-2+%5C+%5C+%5C+%5C+%5C+x_2%3D%5Cfrac%7B4%2B8%7D%7B2%7D%3D6)
ОДЗ
![\left \{ {{x^2-4 \geq 0} \atop {8x+5 \geq 0} \right. \left \{ {{(x-2)(x+2) \geq 0} \atop {x \geq -0,625} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5E2-4+%5Cgeq+0%7D+%5Catop+%7B8x%2B5+%5Cgeq+0%7D+%5Cright.+++%5Cleft+%5C%7B+%7B%7B%28x-2%29%28x%2B2%29+%5Cgeq+0%7D+%5Catop+%7Bx+%5Cgeq+-0%2C625%7D+%5Cright.++)
x∈[2; +
![\infty](https://tex.z-dn.net/?f=%5Cinfty)
)
![\sqrt{x^2-4}-\sqrt{8x+5}=0 \\ \sqrt{x^2-4}=\sqrt{8x+5} \\ (\sqrt{x^2-4})^2=(\sqrt{8x+5})^2 \\ x^2-4=8x+5 \\ x^2-8x-9=0 \\ D=64+36=100 \\ x_1=\frac{8-10}{2}=-1 \ \ \ \ \ \ x_2=\frac{8+10}{2}=9 ](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-4%7D-%5Csqrt%7B8x%2B5%7D%3D0+%5C%5C+%5Csqrt%7Bx%5E2-4%7D%3D%5Csqrt%7B8x%2B5%7D+%5C%5C+%28%5Csqrt%7Bx%5E2-4%7D%29%5E2%3D%28%5Csqrt%7B8x%2B5%7D%29%5E2+%5C%5C+x%5E2-4%3D8x%2B5+%5C%5C+x%5E2-8x-9%3D0+%5C%5C+D%3D64%2B36%3D100+%5C%5C+x_1%3D%5Cfrac%7B8-10%7D%7B2%7D%3D-1+%5C+%5C+%5C+%5C+%5C+%5C+x_2%3D%5Cfrac%7B8%2B10%7D%7B2%7D%3D9%0A)
x₁=-1 не удовлетворяет ОДЗ
Можно решать способом проверки корней
![\sqrt{2x^2-5x+1}=x-1 \\ (\sqrt{2x^2-5x+1})^2=(x-1 )^2 \\ 2x^2-5x+1= x^{2} -2x+1 \\ x^2-3x=0 \\ x(x-3)=0 \\ x=0 \ \ \ \ \ x-3=0 \\ . \ \ \ \ \ \ \ \ \ \ \ \ x=3](https://tex.z-dn.net/?f=%5Csqrt%7B2x%5E2-5x%2B1%7D%3Dx-1+%5C%5C+%28%5Csqrt%7B2x%5E2-5x%2B1%7D%29%5E2%3D%28x-1+%29%5E2+%5C%5C+2x%5E2-5x%2B1%3D+x%5E%7B2%7D+-2x%2B1+%5C%5C+x%5E2-3x%3D0+%5C%5C+x%28x-3%29%3D0+%5C%5C+x%3D0+%5C+%5C+%5C+%5C+%5C+x-3%3D0+%5C%5C++.+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C++%5C+x%3D3)
проверим корни уравнения
х=0
![\sqrt{2*0^2-5*0+1}=0-1 \\ 1=-1](https://tex.z-dn.net/?f=%5Csqrt%7B2%2A0%5E2-5%2A0%2B1%7D%3D0-1+%5C%5C+1%3D-1)
значит х=0 посторонний корень
х=3
![\sqrt{2*3^2-5*3+1}=3-1 \\ 2=2](https://tex.z-dn.net/?f=%5Csqrt%7B2%2A3%5E2-5%2A3%2B1%7D%3D3-1+%5C%5C+2%3D2)
ответ 2