Алгебра 8 класс, номер 1015

Question 1

Question 2

А) $\left \{ {{10,9 - 2x\ \textless \ 0} \atop {12x-90\ \textless \ x}} \right.$

$\left \{ {{2x\ \textgreater \ 10,9} \atop {12x-x\ \textless \ -90}} \right.$

$\left \{ {{x\ \textgreater \ 5,45} \atop {11x\ \textless \ 90}} \right.$

$\left \{ {{x\ \textgreater \ 5,45} \atop {11x\ \textless \ 8 \frac{2}{11}}} \right.$

x∈ (5,45 ; $8 \frac{2}{11}$ ). Целые решения: 6, 7, 8

б) $\left \{ {{1-3x\ \textless \ 3} \atop {6-2x \geq 0}} \right.$

$\left \{ {{-3x\ \textless \ 2} \atop {-2x \geq -6}} \right.$

$\left \{ {{x\ \textgreater \ - \frac{2}{3}} \atop {x \leq 3}} \right.$

x∈ ( $-\frac{2}{3}$ ; 3]. Целые решения: 0, 1, 2, 3

в) $\left \{ {{(x+3)^3\ \textgreater \ x^2(x+9)} \atop {37-12x\ \textgreater \ 0}} \right.$

$\left \{ {{x^3+9x^2+27x+27-x^3-9x^2\ \textgreater \ 0} \atop {37-12x\ \textgreater \ 0}} \right.$

$\left \{ {{27x\ \textgreater \ -27} \atop {-12x\ \textgreater \ -37}} \right.$

$\left \{ {{x\ \textgreater \ -1} \atop {x\ \textless \ 3 \frac{1}{12}}} \right.$

x∈ (-1 ; $3\frac{1}{12}$ ). Целые решения: 0, 1, 2, 3

г) $\left \{ {{(4-x)^3-x^2(12-x)\ \textless \ 0} \atop {7-3x\ \textgreater \ 0}} \right.$

$\left \{ {{(64-3*16*x+3*4*x^2-x^3) -12x^2+x^3\ \textless \ 0} \atop {7-3x\ \textgreater \ 0}} \right.$

$\left \{ {{64-48x+12x^2-x^3 -12x^2+x^3\ \textless \ 0} \atop {7-3x\ \textgreater \ 0}} \right.$

$\left \{ {{-48x\ \textless \ -64} \atop {-3x\ \textgreater \ -7}} \right.$

$\left \{ {{x\ \textgreater \ 1 \frac{1}{3}} \atop {x\ \textless \ 2 \frac{1}{3} }} \right.$

x∈ (1 \frac{1}{3}} ; 2 \frac{1}{3}). Целое решение: 2.