Question

Даю 30 баллов. Алгебра 8 класс. Задания а и б на картинке. Надо найти значения выражений:

Answer 1

$\displaystyle\bf\\\frac{3}{\sqrt{1}+\sqrt{4}} +\frac{3}{\sqrt{4}+\sqrt{7}} +\frac{3}{\sqrt{7}+\sqrt{10}} +...+\frac{3}{\sqrt{97}+\sqrt{100}} =\\\\\\=\frac{3\cdot(\sqrt{1} -\sqrt{4}) }{(\sqrt{1}+\sqrt{4})\cdot(\sqrt{1} -\sqrt{4} )} +\frac{3\cdot(\sqrt{4} -\sqrt{7} )}{(\sqrt{4}+\sqrt{7})\cdot(\sqrt{4} -\sqrt{7} )} +\frac{3\cdot(\sqrt{7} -\sqrt{10} )}{(\sqrt{7}+\sqrt{10})\cdot(\sqrt{7} -\sqrt{10} )} +\\\\\\+...+\frac{3\cdot(\sqrt{97} -\sqrt{100}) }{(\sqrt{97}+\sqrt{100})\cdot(\sqrt{97} -\sqrt{100} )} =$

$\displaystyle\bf\\=\frac{3\cdot(\sqrt{1} -\sqrt{4}) }{1-4} +\frac{3\cdot(\sqrt{4} -\sqrt{7} )}{4-7} +\frac{3\cdot(\sqrt{7} -\sqrt{10} )}{7-10} +...+\frac{3\cdot(\sqrt{97}-\sqrt{100}}{97-100)} =\\\\\\=\frac{3\cdot(\sqrt{1} -\sqrt{4}) }{-3} +\frac{3\cdot(\sqrt{4} -\sqrt{7} )}{-3} +\frac{3\cdot(\sqrt{7} -\sqrt{10} )}{-3} +...+\frac{3\cdot(\sqrt{97}-\sqrt{100})}{-3} =\\\\\\=\sqrt{4}-\sqrt{1} +\sqrt{7} -\sqrt{4 }+\sqrt{10} -\sqrt{7}+...+ \sqrt{100} -\sqrt{97} =\\\\\\=-\sqrt{1} +\sqrt{100} =-1+10=9$

$\displaystyle\bf\\2)\\\\\frac{3}{\sqrt{9}+\sqrt{12}} +\frac{3}{\sqrt{12}+\sqrt{15}} +\frac{3}{\sqrt{15}+\sqrt{18}} +...+\frac{3}{\sqrt{118}+\sqrt{121}} =\\\\\\=\frac{3\cdot(\sqrt{9} -\sqrt{12}) }{(\sqrt{9}+\sqrt{12})\cdot(\sqrt{9} -\sqrt{12} )} +\frac{3\cdot(\sqrt{12} -\sqrt{15} )}{(\sqrt{12}+\sqrt{15})\cdot(\sqrt{12} -\sqrt{15} )} +\frac{3\cdot(\sqrt{15} -\sqrt{18} )}{(\sqrt{15}+\sqrt{18})\cdot(\sqrt{15} -\sqrt{18} )} +\\\\\\+$$\displaystyle\bf\+......+\frac{3\cdot(\sqrt{118} -\sqrt{121}) }{(\sqrt{118}+\sqrt{121})\cdot(\sqrt{118} -\sqrt{121} )} =\\\\\\=$

$\displaystyle\bf\\=\frac{3\cdot(\sqrt{9} -\sqrt{12}) }{9-12} +\frac{3\cdot(\sqrt{12} -\sqrt{15} )}{12-15} +\frac{3\cdot(\sqrt{15} -\sqrt{18} )}{15-18} +...+\frac{3\cdot(\sqrt{118}-\sqrt{121}}{118-121} =\\\\\\=\frac{3\cdot(\sqrt{9} -\sqrt{12}) }{-3} +\frac{3\cdot(\sqrt{12} -\sqrt{15} )}{-3} +\frac{3\cdot(\sqrt{15} -\sqrt{18} )}{-3} +...+\frac{3\cdot(\sqrt{118}-\sqrt{121})}{-3} =$

$\displaystyle\bf\\=\sqrt{12}-\sqrt{9} +\sqrt{15} -\sqrt{12 }+\sqrt{18} -\sqrt{15}+...+ \sqrt{121} -\sqrt{118} =\\\\\\=-\sqrt{9} +\sqrt{121} =-3+11=8$