Y=3x²-4x
a=3>0 ⇒ ветви вверх
х(верш)= -( -4)/(2·3)=4/6=2/3
у(верш)=3·(4/9) -4·(2/3)=4/3-8/3= -4/3 - наименьшее значение у
3х+4= 2/3-2
Умножим все уравнение на 3, что бы избавится от дроби.
9х+12=2-6
9х=2-6-12
9х= -16
х = -16/9
х= -1 7/9 (одна целая и семь девятых)
Во всех примерах выражение не имеет смысла при знаменателе равном нулю
![a)y-8=0\ \ \ \ \ \ \ b)8g-3=0\ \ \ \ \ \ \ \ \ c)7-19z=0\\y=8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g=\frac{3}{8}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ z=\frac{7}{19}\\\\d)12p-5=0\ \ \ \ e)-14-3u=0\ \ \ \ f)2x-6=0\\p=\frac{5}{12}\ \ \ \ \ \ \ \ \ \ \ \ \ u=-\frac{3}{14}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=3\\\\g)13b-4=0\ \ \ \ \ h)2a+9=0\\b=\frac{4}{13}\ \ \ \ \ \ \ \ \ \ \ \ \ \ a=-4,5](https://tex.z-dn.net/?f=a%29y-8%3D0%5C+%5C+%5C+%5C+%5C+%5C+%5C+b%298g-3%3D0%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+c%297-19z%3D0%5C%5Cy%3D8%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+g%3D%5Cfrac%7B3%7D%7B8%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C++z%3D%5Cfrac%7B7%7D%7B19%7D%5C%5C%5C%5Cd%2912p-5%3D0%5C+%5C+%5C+%5C+e%29-14-3u%3D0%5C+%5C+%5C+%5C+f%292x-6%3D0%5C%5Cp%3D%5Cfrac%7B5%7D%7B12%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+u%3D-%5Cfrac%7B3%7D%7B14%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+x%3D3%5C%5C%5C%5Cg%2913b-4%3D0%5C+%5C+%5C+%5C+%5C+h%292a%2B9%3D0%5C%5Cb%3D%5Cfrac%7B4%7D%7B13%7D%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+%5C+a%3D-4%2C5)