Question

Решите уравнение
(4а²-1)/3=(10а-9)а

Дам 50 баллов

Answer 1

$\frac{4a^{2}-1 }{3}=\frac{10a-9}{a}\\\\\frac{4a^{2}-1 }{3}-\frac{10a-9}{a} =0\\\\\frac{4a^{3}-a-30a+27 }{3a}=0\\\\\frac{4a^{3}-31a+27 }{3a}=0\\\\\frac{(4a^{3}-4a)-(27a-27)}{3a}=0\\\\\frac{4a(a^{2}-1)-27(a-1)}{3a}=0\\\\\frac{(a-1)(4a^{2}+4a-27) }{3a}=0\\\\\left \{ {{(a-1)(4a^{2}+4a-27)=0 } \atop {a\neq0 }} \right.\\\\1)a-1=0\\\\a_{1} =1\\\\2)4a^{2} +4a-27=0\\\\D=4^{2}-4*4*(-27)=16+432=448=(8\sqrt{7})^{2} \\\\a_{2} =\frac{-4-8\sqrt{7}}{8} =-\frac{1+2\sqrt{7}}{2}$

$a_{3} =\frac{-4+8\sqrt{7}}{8} =\frac{2\sqrt{7}-1}{2}\\\\Otvet:\boxed{1;-\frac{1+2\sqrt{7}}{2};\frac{2\sqrt{7}-1}{2}}$

Если уравнение выглядит так :

$\frac{4a^{2}-1 }{3}=(10a-9)a$

То решение такое :

$\frac{4a^{2}-1 }{3}*3=(10a-9)a*3\\\\4a^{2}-1=30a^{2} -27a\\\\30a^{2}-4a^{2}-27a+1=0\\\\26a^{2}-27a+1=0\\\\D=(-27)^{2} -4*26*1=729-104=625=25^{2}\\\\a_{1}=\frac{27-25}{52}=\frac{2}{52}=\frac{1}{26}\\\\a_{2}=\frac{27+25}{52}=\frac{52}{52}=1\\\\Otvet:\boxed{\frac{1}{26};1}$