Question

помогите справиться с алгеброй.Надо само решение.​

Answer 1

Ответ:

$1)\ \ 3(x^2-36)=0\\\\(x-6)(x+6)=0\\\\x-6=0\ \ \to \ \ x_1=6\\\\x+6=0\ \ \to \ \ x_2=-6\\\\Otvet:\ \ \underline {\ x_1=6\ ,\ x_2=-6\ }\ .\\\\\\2)\ \ x^2-144=0\\\\(x-12)(x+12)=0\\\\\underline {\ x_1=12\ ,\ \ x_2=-12\ }$

$3)\ \ \dfrac{19-x^2}{15}=1\ \ \ \to \ \ \ \dfrac{19-x^2-15}{15}=0\ \ \ ,\ \ \ \dfrac{4-x^2}{15}=0\ \ ,\ \ \ 4-x^2=0\\\\(2-x)(2+x)=0\\\\\underline {\ x_1=2\ ,\ \ x_2=-2\ }\\\\\\4)\ \ x^2=100\ \ \ \to \ \ \ x^2-100=0\\\\(x-10)(x+10)=0\\\\\underline {\ x_1=10\ \ ,\ \ x_2=-10\ }$