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Ответ:

2.1 [tex](3\sqrt{6} +2\sqrt{8} -\sqrt{32})\sqrt{2} -\sqrt{108}[/tex]

[tex](3\sqrt{6}+4\sqrt{2} -4\sqrt{2} )\sqrt{2} -6\sqrt{3}[/tex])

[tex]3\sqrt{6} \sqrt{2} -6\sqrt{3}[/tex]

[tex]3\sqrt{12} -6\sqrt{3}[/tex]

[tex]6\sqrt{3} -6\sqrt{3}[/tex]

0

2.2)методом подстановки

[tex]\left \{ {{x+y=2} \atop {2x^{2}-xy=65 }} \right.[/tex]

[tex]\left \{ {{x=2-y} \atop {2x^{2}-xy=65 }} \right.[/tex]

2(2y)^2-(2-y)y=65

y=-3

y=19/3

x=2-(-3)

x=2-19/3

x=5

x=-13/3

(5;-3)

(-13/3;19/3)

2.3 0.1(7)=17-1/90=16/90=8/45

2.4 [tex]\frac{5b}{b-3} -\frac{b+6}{2b-6} *\frac{90}{b^{2} +6b}[/tex]

[tex]\frac{5b}{b-3} -\frac{b+6}{2(b-6)} *\frac{90}{b(b+6)}[/tex]

[tex]\frac{5b}{b-3} -\frac{45}{b(b-3)} *\frac{45}{b}[/tex]

[tex]\frac{5b}{b-3} -\frac{45}{b(b-3)}[/tex]

[tex]\frac{5b^{2}-45 }{b(b-3)}[/tex]

[tex]\frac{5(b-3(b+3)}{b(b-3)}[/tex]

[tex]\frac{5(b+3)}{b}[/tex]

[tex]\frac{5b+15}{b}[/tex]

Объяснение: