10(1138). For what value of $a$ do the graphs of the functions $y=x^{2}-7 x+a$ and $y=-3 x^{2}+5 x-6$ have exactly one common point? Find its coordinates.
Solution. For the graphs of the given functions to have a common point, it is necessary that the equation $x^{2}-7 x+a = -3 x^{2}+5 x-6$ holds, from which we get $4 x^{2}-12 x+(a+6)=0$.

For the obtained quadratic trinomial to have one root (two equal roots), and for the graphs of the given functions to have a unique common point, it is necessary that the discriminant of the quadratic trinomial $4 x^{2}-12 x+(a+6)$ be equal to zero. We have $D=144-16(a+6)=0$, from which $a=3$.

Substituting $a=3$ into the equation $4 x^{2}-12 x+(a+6)=0$, we get the equation $4 x^{2}-12 x+9=0$, which has a unique root $x=1.5$. Therefore, the graphs of the given functions intersect at one point, the abscissa of which is 1.5. Substituting $x=1.5$ into the formula of the given function $y=x^{2}-7 x+3$ or into the formula $y=-3 x^{2}+5 x-6$, we find the ordinate of the desired point: $y=-5.25$.

Answer: $a=3, x=1.5, y=-5.25$.

$\triangle 11(1139)$. Solve the inequality:
a) $\frac{3 x^{2}+4 x-4}{x^{2}+x+1} > 1$.

Solution. a) $\frac{3 x^{2}+4 x-4}{x^{2}+x+1} - 1 > 0$ for all real values of $x$, then $2 x^{2}+3 x-5 > 0$, $\frac{3 x^{2}+2 x-1-x^{2}+x-1}{x^{2}-x+1} > 0$,

$$
\frac{2 x^{2}+3 x-2}{x^{2}-x+1} > 0, \quad \frac{2(x+2)\left(x-\frac{1}{2}\right)}{x^{2}-x+1} > 0
$$

Since $x^{2}-x+1 > 0$, then $2(x+2)\left(x-\frac{1}{2}\right) > 0$, from which $x < -2$ or $x > \frac{1}{2}$.

Answer: a) $x < -2$ or $x > \frac{1}{2}$.