Можно применить формулу разности квадратов $(a-b)(a+b)=a^2-b^2$:

$(5+\sqrt7)(5-\sqrt7)=5^2-(\sqrt7)^2=25-7=18$.

$\left({\Large\frac{a}{b^3}}\right)^6\cdot\left({\Large\frac{b^{10}}{a^3}}\right)^2={\Large\frac{a^{6}}{b^{3\cdot6}}}\cdot{\Large\frac{b^{10\cdot2}}{a^{3\cdot2}}}={\Large\frac{a^6\cdot b^{20}}{b^{18}a^6}}=b^2$ при $b=6$

$6^2=36$

Ответ: $36$.

$(a-b)^2=a^2-2ab+b^2$

$(5a-2b)^2+20ab=(5a)^2-2\cdot5a\cdot2b+(2b)^2+20ab=25a^2-20ab+4b^2+20ab=25a^2+4b^2$

при $a=\sqrt2$, $b=\sqrt3$

$25\cdot(\sqrt2)^2+4\cdot(\sqrt3)^2=25\cdot2+4\cdot3=50+12=62$

Ответ: $62$.

${\Large\frac{y^2-4y+4}{y^2-2y}}={\Large\frac{(y-2)^2}{y(y-2)}}={\Large\frac{y-2}{y}}$

При $y=0,2$

${\Large\frac{0,2-2}{0,2}}={\Large\frac{-1,8}{0,2}}=-{\Large\frac{18}{2}}=-9$

Ответ: $-9$.

${\Large\frac{x-3}{x^2}}:{\Large\frac{2x-6}{x^2+3x}}={\Large\frac{(x-3)\cdot x\cdot (x+3)}{x^2\cdot2\cdot(x-3)}}={\Large\frac{1\cdot1\cdot(x+3)}{x\cdot2\cdot1}}={\Large\frac{x+3}{2x}}$

При $x={\Large\frac{1}{2}}$

${\Large\frac{\frac{1}{2}+3}{2\cdot\frac{1}{2}}}={\Large\frac{3,5}{1}}=3,5$

Ответ: $3,5$.

${\Large\frac{6y}{y-x}}+\left({\Large\frac{2xy}{x^2-y^2}}+{\Large\frac{x-y}{2x+2y}}\right)\cdot{\Large\frac{12x}{x+y}}$

1. ${\Large\frac{2xy}{x^2-y^2}}+{\Large\frac{x-y}{2x+2y}}={\Large\frac{2xy}{(x-y)(x+y)}}+{\Large\frac{x-y}{2(x+y)}}={\Large\frac{2\cdot2xy+(x-y)(x-y)}{2(x-y)(x+y)}}={\Large\frac{4xy+x^2-2xy+y^2}{2(x-y)(x+y)}}={\Large\frac{x^2+2xy+y^2}{2(x-y)(x+y)}}={\Large\frac{(x+y)^2}{2(x-y)(x+y)}}={\Large\frac{x+y}{2(x-y)}}$ 2. ${\Large\frac{x+y}{2(x-y)}}\cdot{\Large\frac{12x}{x+y}}={\Large\frac{6x}{x-y}}$ 3. ${\Large\frac{6y}{y-x}}+{\Large\frac{6x}{x-y}}={\Large\frac{6y}{y-x}}+{\Large\frac{6x}{-(-x+y)}}={\Large\frac{6y}{y-x}}-{\Large\frac{6x}{y-x}}={\Large\frac{6y-6x}{y-x}}={\Large\frac{6(y-x)}{y-x}}=6$

Ответ: $6$.