Question

Б Упростите выражения (141-142): 141. (10-7√√2)²+(10+7√√2)²; 2) (1+ √2-√3)(1+√2 + √3); 3(1-√2-√3)(1+√√2 +√3); (√6-√5)(√30-√6-√5); 5) (₁ √5 + 3√² ) - (3√/2² - 1 √5): 6 (5 + /3 - /15)(5 - √3) + √75 ; 7) (; √39 - } √26 + { √65); ¿ √I3 + √18; 8)/45-110 (33 + 1,5/22 - 0,S√55). 2 6
дам 50 балов
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Answer 1

Решение.

Упростить выражения .

$\bf 1)\ \ (10-7\sqrt2)^2+(10+7\sqrt2)^2=\\\\=100-20\cdot 7\sqrt2+49\cdot 2+(100+20\cdot 7\sqrt2+49\cdot 2)=200+196=396\\\\\\2)\ \ (1+\sqrt2-\sqrt3)(1+\sqrt2+\sqrt3)=(1+\sqrt2)^2-(\sqrt3)^2=3+2\sqrt2-3=2\sqrt2\\\\\\3)\ \ (1-\sqrt2-\sqrt3)(1+\sqrt2+\sqrt3)=1^2-(\sqrt2+\sqrt3)^2=1-(2+3+2\sqrt6)=\\\\=-4-2\sqrt6=-2(2+\sqrt6)\\\\\\4)\ \ (\sqrt6-\sqrt5)(\sqrt{30}-\sqrt6-\sqrt5)=(\sqrt6-\sqrt5)\cdot \sqrt{30}-(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)=\\\\=\sqrt{180}-\sqrt{150}-(6-5)=6\sqrt5-5\sqrt6-1$

$\bf 5)\ \ (\frac{1}{3}\sqrt5+3\sqrt2)^2-(3\sqrt2-\frac{1}{3}\sqrt5)^2=\\\\=\dfrac{5}{9}+9\cdot 2+\dfrac{2}{3}\cdot \sqrt5\cdot 3\sqrt2-\Big(\dfrac{5}{9}+9\cdot 2-\dfrac{2}{3}\cdot \sqrt5\cdot 3\sqrt2\Big)=\dfrac{4}{3}\cdot \sqrt5\cdot 3\sqrt2=4\sqrt{10}\\\\\\6)\ \ (\sqrt5+\sqrt3-\sqrt{15})(\sqrt5-\sqrt3)+\sqrt{75}=\\\\=(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)-\sqrt{15}\, (\sqrt5-\sqrt3)+\sqrt{75}=\\\\=(5-3)-\sqrt{75}+\sqrt{45}+\sqrt{75}=2+\sqrt{45}=2+3\sqrt5$  

$\displaystyle \bf 7)\ \ \Big(\dfrac{1}{3}\sqrt{39}-\dfrac{1}{2}\sqrt{26}+\dfrac{1}{6}\sqrt{65}\Big):\dfrac{1}{6}\sqrt{13}+\sqrt{18}=\\\\\\=\dfrac{\sqrt{3\cdot 13}}{9}\cdot \dfrac{6}{\sqrt{13}}}-\frac{\sqrt{2\cdot 13}}{2}\cdot \frac{6}{\sqrt{13}}+\frac{\sqrt{5\cdot 13}}{6}\cdot \frac{6}{\sqrt{13}}+\sqrt{18}=\\\\\\=\frac{2\sqrt3}{3}-3\sqrt2+\sqrt5+3\sqrt{2}=\frac{2\sqrt3}{3}+\sqrt5=\frac{2\sqrt3+3\sqrt5}{3}$

$\displaystyle \bf 8)\ \ \sqrt{45}-\frac{1}{11}\sqrt{110}\cdot (\sqrt{33}+1,5\sqrt{22}-0,5\sqrt{55})=\\\\\\=3\sqrt5-\frac{\sqrt{11\cdot 10}}{11}\cdot \sqrt{11\cdot 3}-\frac{3\sqrt{11\cdot 10}}{2\cdot 11}\sqrt{11\cdot 2}+\frac{\sqrt{11\cdot 10}}{11\cdot 2}\cdot \sqrt{11\cdot 5}=\\\\\\=3\sqrt5-\sqrt{10\cdot 3}-\frac{3\sqrt{10\cdot 2}}{2}+\frac{\sqrt{10\cdot 5}}{2}=\\\\\\=3\sqrt5-\sqrt{30}-3\sqrt5+\frac{5\sqrt2}{2}=\frac{5\sqrt2}{2}-\sqrt{30}$