Question

Выражение 10a^-1/b^-2-a^-2+5/b^-1+a^-1 и найдите его значение при a=4/9,b=2/3

Answer 1

$a=\frac{4}{9}\; ,\; \; b=\frac{2}{3}\\\\\frac{10a^{-1}}{b^{-2}}-\frac{a^{-2}+5}{b^{-1}+a^{-1}}=\frac{10b^2}{a}-\frac{(1+5a^2)\cdot ab}{a^2\cdot (a+b)}=\frac{10b^2(a+b)-b(1+5a^2)}{a(a+b)}=\\\\=\frac{b(10ab+10b^2-1-5a^2)}{a(a+b)}=\frac{\frac{2}{3}(\frac{80}{27}+\frac{40}{9}-1-\frac{80}{81})}{\frac{4}{9}\cdot (\frac{4}{9}+\frac{2}{3})}=\frac{3\cdot \frac{439}{81}}{2\cdot \frac{10}{9}}=\frac{1317}{180}=7\frac{57}{180}$

$2)\; \frac{10a^{-1}}{b^{-2}-a^{-2}}-\frac{5}{b^{-1}+a^{-1}}=\frac{10}{a\cdot (\frac{1}{b^2}-\frac{1}{a^2})}-\frac{5}{\frac{1}{b}+\frac{1}{a}}=\frac{10\cdot a^2b^2}{a\cdot (a-b)(a+b)}-\frac{5\cdot ab}{a+b}=\\\\=\frac{10ab^2-5ab\cdot (a-b)}{(a-b)(a+b)}=\frac{5ab\cdot (2b-a+b)}{(a-b)(a+b)}=\frac{5ab\cdot (3b-a)}{(a-b)(a+b)}=-9\frac{1}{3}\\\\5ab(3b-a)=5\cdot \frac{4\cdot 2}{9\cdot 3}\cdot (\frac{6}{3}-\frac{4}{9})=\frac{5\cdot 4\cdot 2}{9\cdot 3}\cdot \frac{14}{9}=\frac{560}{243}\\\\(a-b)(a+b)=(\frac{4}{9}-\frac{2}{3})(\frac{4}{9}+\frac{2}{3})=-\frac{2}{9}\cdot \frac{10}{9}=-\frac{20}{81}$

$\frac{560/243}{-20/81}=-\frac{560\cdot 81}{243\cdot 20} =-\frac{28}{3}=-9\frac{1}{3}$

Answer 2

10a^-1/b^-2-a^-2+5/b^-1+a^-1=10*(4/9)^-1/((2/3)^-2-(4/9)^-2)+5/(2/3)^-1+(4/9)-1=((90/4)/-45/16)+5/(15/4)=-20/3