При каких значениях а дробь а^3-25/ а^2-4a+5 равна нулю
Дробь равна нулю, если её числитель равен нулю, знаменатель не равен нулю. Ответ: при а = ∛25
Не то
в каком смысле?
![\left \{ {{a^3-25=0} \atop {a^2-4a+5 \neq 0}} \right. \ \ \to \left \{ {{a^3=25} \atop {(a-2)^2+1 \neq 0}} \right. \ \ \to \left \{ {{a= \sqrt[3]{25} } \atop {(a-2)^2 \neq 1}} \right. \ \ \to \left \{ {{a= \sqrt[3]{25} } \atop {a-2\neq б1}} \right. \ \ \to \\\\\\
\left \{ {{a= \sqrt[3]{25} } \atop {a \neq1; \ a \neq 3 }} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Ba%5E3-25%3D0%7D+%5Catop+%7Ba%5E2-4a%2B5+%5Cneq+0%7D%7D+%5Cright.++%5C+%5C+%5Cto++%5Cleft+%5C%7B+%7B%7Ba%5E3%3D25%7D+%5Catop+%7B%28a-2%29%5E2%2B1+%5Cneq+0%7D%7D+%5Cright.++%5C+%5C+%5Cto+%5Cleft+%5C%7B+%7B%7Ba%3D+%5Csqrt%5B3%5D%7B25%7D+%7D+%5Catop+%7B%28a-2%29%5E2+%5Cneq+1%7D%7D+%5Cright.++%5C+%5C+%5Cto+%5Cleft+%5C%7B+%7B%7Ba%3D+%5Csqrt%5B3%5D%7B25%7D+%7D+%5Catop+%7Ba-2%5Cneq+%D0%B11%7D%7D+%5Cright.++%5C+%5C+%5Cto+%5C%5C%5C%5C%5C%5C+%0A+%5Cleft+%5C%7B+%7B%7Ba%3D+%5Csqrt%5B3%5D%7B25%7D+%7D+%5Catop+%7Ba+%5Cneq1%3B++%5C+a+%5Cneq+3+%7D%7D+%5Cright. "\left \{ {{a^3-25=0} \atop {a^2-4a+5 \neq 0}} \right. \ \ \to \left \{ {{a^3=25} \atop {(a-2)^2+1 \neq 0}} \right. \ \ \to \left \{ {{a= \sqrt[3]{25} } \atop {(a-2)^2 \neq 1}} \right. \ \ \to \left \{ {{a= \sqrt[3]{25} } \atop {a-2\neq б1}} \right. \ \ \to \\\\\\
\left \{ {{a= \sqrt[3]{25} } \atop {a \neq1; \ a \neq 3 }} \right.")