Question

решите уравнение неполным способом
6у^2+9y=0
16x^2-1=0

Answer 1

1)

$\displaystyle \tt 6y^2+9y=0\\\displaystyle \tt 3y(2y+3)=0\\\displaystyle \tt y(2y+3)=0\\\\ \displaystyle \tt \bold{y_1=0}\\\\ \displaystyle \tt 2y+3=0\\\displaystyle \tt 2y=-3\\\displaystyle \tt y=-3\div2\\\displaystyle \tt \bold{y_2=-1,5}$

2)

$\displaystyle \tt 16x^2-1=0\\\\ \displaystyle \tt 16x^2=1\\\\\displaystyle \tt x^2=\frac{1}{16}\\\\ \displaystyle \tt x=б\sqrt{\frac{1}{16}}\\\\ \displaystyle \tt \bold{x_1=\frac{1}{4}}\\\\ \displaystyle \tt \bold{x_2=-\frac{1}{4}}$

Answer 2

Ответ:

$6y^{2} +9y=0\\3y(2y+3)=0\\y(2y+3)=0\\y=0\\2y+3=0$⇒$y=-\frac{3}{2} =-1.5$

y=-1.5

y=0

$16x^{2} -1=0\\16x^{2} =1\\x^{2} =\frac{1}{16}\\ x=+-\frac{1}{4}$

x=-1/4

x=1/4