Упр.821 ГДЗ Никольский Потапов 9 класс (Алгебра)

б) (1/2009^2-1/2010^2):(1/2009+1/2010)·2009^2;
в) (1/2010^2-1/2011^2):(1/2010-1/2011)·2011/4021.

1. а)

   $$\left(\frac{1}{49^2}-\frac{1}{50^2}\right):\left(\frac{1}{49}-\frac{1}{50}\right)\cdot\frac{7}{9}$$
   $$=\left(\frac{1}{49}-\frac{1}{50}\right)\left(\frac{1}{49}+\frac{1}{50}\right):\left(\frac{1}{49}-\frac{1}{50}\right)\cdot\frac{7}{9}$$
   $$=\left(\frac{1}{49}+\frac{1}{50}\right)\cdot\frac{7}{9}$$
   $$=\frac{50+49}{49\cdot 50}\cdot\frac{7}{9}=\frac{99}{49\cdot 50}\cdot\frac{7}{9}=\frac{11}{350}.$$

2. б)

   $$\left(\frac{1}{2009^2}-\frac{1}{2010^2}\right):\left(\frac{1}{2009}+\frac{1}{2010}\right)\cdot 2009^2$$
   $$=\left(\frac{1}{2009}-\frac{1}{2010}\right)\left(\frac{1}{2009}+\frac{1}{2010}\right):\left(\frac{1}{2009}+\frac{1}{2010}\right)\cdot 2009^2$$
   $$=\left(\frac{1}{2009}-\frac{1}{2010}\right)\cdot 2009^2$$
   $$=\frac{2010-2009}{2009\cdot 2010}\cdot 2009^2=\frac{2009}{2010}.$$

3. в)

   $$\left(\frac{1}{2010^2}-\frac{1}{2011^2}\right):\left(\frac{1}{2010}-\frac{1}{2011}\right)\cdot\frac{2011}{4021}$$
   $$=\left(\frac{1}{2010}-\frac{1}{2011}\right)\left(\frac{1}{2010}+\frac{1}{2011}\right):\left(\frac{1}{2010}-\frac{1}{2011}\right)\cdot\frac{2011}{4021}$$
   $$=\left(\frac{1}{2010}+\frac{1}{2011}\right)\cdot\frac{2011}{4021}$$
   $$=\frac{2010+2011}{2010\cdot 2011}\cdot\frac{2011}{4021}=\frac{4021}{2010\cdot 2011}\cdot\frac{2011}{4021}=\frac{1}{2010}.$$

Ответ

а) $$\frac{11}{350}$$; б) $$\frac{2009}{2010}$$; в) $$\frac{1}{2010}$$.