Упр.645 ГДЗ Никольский Потапов 8 класс (Алгебра)

• а) $$\left(2\sqrt{\frac{3}{5}}+\sqrt{\frac{3}{8}}\right)\left(\sqrt{\frac{3}{8}}-2\sqrt{\frac{3}{5}}\right)$$; б) $$\left(3\sqrt{\frac{5}{6}}-\sqrt{\frac{3}{5}}\right)\left(3\sqrt{\frac{5}{6}}+\sqrt{\frac{3}{5}}\right)$$; в) $$\left(\sqrt{13+5\sqrt{4{,}2}}+\sqrt{13-5\sqrt{4{,}2}}\right)^2$$; г) $$\left(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\right)^2$$.

1. а)

   $$\left(2\sqrt{\frac{3}{5}}+\sqrt{\frac{3}{8}}\right)\left(\sqrt{\frac{3}{8}}-2\sqrt{\frac{3}{5}}\right) =\left(\sqrt{\frac{3}{8}}\right)^2-\left(2\sqrt{\frac{3}{5}}\right)^2$$
   $$=\frac{3}{8}-4\cdot\frac{3}{5} =\frac{3}{8}-\frac{12}{5} =\frac{15-96}{40} =-\frac{81}{40} =-2\frac{1}{40}.$$

2. б)

   $$\left(3\sqrt{\frac{5}{6}}-\sqrt{\frac{3}{5}}\right)\left(3\sqrt{\frac{5}{6}}+\sqrt{\frac{3}{5}}\right) =\left(3\sqrt{\frac{5}{6}}\right)^2-\left(\sqrt{\frac{3}{5}}\right)^2$$
   $$=9\cdot\frac{5}{6}-\frac{3}{5} =\frac{75}{10}-\frac{6}{10} =\frac{69}{10} =6{,}9.$$

3. в)

   $$\left(\sqrt{13+5\sqrt{4{,}2}}+\sqrt{13-5\sqrt{4{,}2}}\right)^2$$
   $$=\left(\sqrt{13+5\sqrt{4{,}2}}\right)^2+2\sqrt{(13+5\sqrt{4{,}2})(13-5\sqrt{4{,}2})}+\left(\sqrt{13-5\sqrt{4{,}2}}\right)^2$$
   $$=13+5\sqrt{4{,}2}+2\sqrt{169-25\cdot 4{,}2}+13-5\sqrt{4{,}2}$$
   $$=26+2\sqrt{169-105} =26+2\sqrt{64} =26+2\cdot 8 =42.$$

4. г)

   $$\left(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\right)^2$$
   $$=\left(\sqrt{11+6\sqrt{2}}\right)^2-2\sqrt{(11+6\sqrt{2})(11-6\sqrt{2})}+\left(\sqrt{11-6\sqrt{2}}\right)^2$$
   $$=11+6\sqrt{2}-2\sqrt{121-36\cdot 2}+11-6\sqrt{2}$$
   $$=22-2\sqrt{121-72} =22-2\sqrt{49} =22-2\cdot 7 =8.$$

Ответ

а) $$-\frac{81}{40}$$; б) $$\frac{69}{10}$$; в) $$42$$; г) $$8$$.