Răspuns
Explicație pas cu pas:
[tex]a=\left(\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\ldots + \dfrac{1}{99\cdot 100}\right)\left(\dfrac{2}{1\cdot 3}+\dfrac{2}{3\cdot 5}+\ldots+\dfrac{2}{97\cdot 99}\right)\\a=\left(\dfrac{2-1}{1\cdot 2}+\dfrac{3-2}{2\cdot 3}+\ldots+\dfrac{100-99}{99\cdot 100}\right)\left(\dfrac{3-1}{1\cdot 3}+\dfrac{5-3}{3\cdot 5}+\ldots+\dfrac{99-97}{97\cdot 99}\right)[/tex]
[tex]a=\left(\dfrac{2}{1\cdot 2}-\dfrac{1}{1\cdot 2}+\dfrac{3}{2\cdot 3}-\dfrac{2}{2\cdot 3}+\ldots +\dfrac{100}{99\cdot 100}-\dfrac{99}{99\cdot 100}\right)\left(\dfrac{3}{1\cdot 3} \right - \\\left \dfrac{1}{1\cdot 3}+\dfrac{5}{3\cdot 5}-\dfrac{3}{3\cdot 5}+\ldots +\dfrac{99}{97\cdot 99}-\dfrac{97}{97\cdot 99}\right)\\a=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\ldots+\dfrac{1}{99}-\dfrac{1}{100}\right)\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\ldots +\dfrac{1}{97}-\dfrac{1}{99}\right)[/tex]
[tex]a=\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)=\dfrac{99}{100}\cdot \dfrac{98}{99} =\dfrac{98}{100}[/tex]
[tex]\sqrt{0,5\cdot a}+0,3= \sqrt{\dfrac{1}{2}\cdot \dfrac{98}{100}}+\dfrac{3}{10}= \sqrt{\dfrac{49}{100}}+\dfrac{3}{10}=\dfrac{7}{10}+\dfrac{3}{10}=\dfrac{7+3}{10}=\dfrac{10}{10}\\=\bold{1}[/tex]
