Question

одно упражнение, алгебра 8 класс, 10 минут​

Answer 1

$3a+\dfrac1{2a}=5\qquad ODZ: \ \ a\ne0\\\\\dfrac{3a\cdot2a+1}{2a}-5=0\\\\\dfrac{6a^2+1-5\cdot2a}{2a}=0\\\\\dfrac{6a^2-10a+1}{2a}\\\\6a^2-10a+1=0\\\\\boxed{D = b^2-4ac}\\\\D=(-10)^2-4\cdot6\cdot1=76$

76 > 0  =>  2 корня

$\sqrt{76}=2\sqrt{19}\\\\\begin{array}{lcl}a_1=\dfrac{10-2\sqrt{19}}{2\cdot6}=\dfrac{5-\sqrt{19}}{6}\\\\a_2=\dfrac{10+2\sqrt{19}}{2\cdot6}=\dfrac{5+\sqrt{19}}{6}\end{array} \qquad\qquad\boxed{x=\dfrac{-b\underline+\sqrt{D}}{2a}}$

$9a^2+\dfrac1{4a^2}\quad \pi pu\quad a=\dfrac{5-\sqrt{19}}{6}$

$1)\ \ 9\bigg(\dfrac{5-\sqrt{19}}6\bigg)^2+\dfrac1{4\bigg(\dfrac{5-\sqrt{19}}{6}\bigg)^2}=9\cdot\dfrac{25-10\sqrt{19}+19}{36}+\dfrac1{4\cdot\dfrac{25-10\sqrt{19}+19}{36}}=$

$=\dfrac{44-10\sqrt{19}}{4}+1:\dfrac{44-10\sqrt{19}}{9}=\dfrac{(44-10\sqrt{19})^2+9\cdot4}{4(44-10\sqrt{19})}=$$\dfrac{44^2-880\sqrt{19}+1900+36}{4(44-10\sqrt{19})}=\dfrac{4(484-220\sqrt{19}+475+9)}{4(44-10\sqrt{19})}=\dfrac{968-220\sqrt{19}}{44-10\sqrt{19}}=$

$=\dfrac{22(44-10\sqrt{19})}{44-10\sqrt{19}}=22$

$9a^2+\dfrac1{4a^2}\quad \pi pu\quad a=\dfrac{5+\sqrt{19}}{6}$

$2)\ \ 9\bigg(\dfrac{5+\sqrt{19}}{6}\bigg)^2+\dfrac1{4\bigg(\dfrac{5+\sqrt{19}}{6}\bigg)^2}=9\cdot\dfrac{25+10\sqrt{19}+19}{36}+\dfrac1{4\cdot\dfrac{25+10\sqrt{19}+19}{36}}=$

$=\dfrac{44+10\sqrt{19}}4+1:\dfrac{44+10\sqrt{19}}9=\dfrac{(44+10\sqrt{19})^2+9\cdot4}{4(44+10\sqrt{19})}=$

$=\dfrac{1936+880\sqrt{19}+1900+36}{4(44+10\sqrt{19})}=\dfrac{3872+880\sqrt{19}}{4(44+10\sqrt{19})}=\dfrac{88(44+10\sqrt{19})}{4(44+10\sqrt{19})}=22$

Ответ:  22