Question

Решите систему неравенств:
5х2-9х + 4 <0,
{
2x + 3 ≥ 0​

Answer 1

Ответ:

$\left\{\begin{array}{l}5x^2-9x+4<0\\2x+3\geq 0\end{array}\right\ \ \left\{\begin{array}{l}5(x-\frac{4}{5})(x-1)<0\\2x\geq -3\end{array}\right\ \ \left\{\begin{array}{l}x\in (\frac{4}{5}\ ;1\ )\\x\in [-1,5\, ;+\infty )\end{array}\right\\\\\\5x^2-9x+4=0\ \ ,\ \ D=81-80=1\ ,\ \ x_1=\dfrac{4}{5}\ ,\ x_2=1\\\\\star \ \ 5x^2-9x+4=5(x-\frac{4}{5})(x-1)\ \ ,\ \ +++(\frac{4}{5})---(1)+++\ \ \star \\\\\boxed{\ x\in \Big(\frac{4}{5}\ ;\ 1\ \Big)\ }\ \ \ ili\ \ \ \boxed{\ x\in \Big(\ 0,8\ ;\ 1\ \Big)\ }$