пусть первое число х второе у тогда
![\displaystyle \left \{ {{x+y=25} \atop {x*y=144}} \right.\\\\ \left \{ {{x=25-y} \atop {(25-y)y=144}} \right.\\\\25y-y^2=144\\\\y^2-25y+144=0\\\\D=625-576=49\\\\y_{1.2}= \frac{25 \pm 7}{2}; y_1=16; y_2=9\\\\x_1=25-16=9; x_2=25-9=16](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cleft+%5C%7B+%7B%7Bx%2By%3D25%7D+%5Catop+%7Bx%2Ay%3D144%7D%7D+%5Cright.%5C%5C%5C%5C+%5Cleft+%5C%7B+%7B%7Bx%3D25-y%7D+%5Catop+%7B%2825-y%29y%3D144%7D%7D+%5Cright.%5C%5C%5C%5C25y-y%5E2%3D144%5C%5C%5C%5Cy%5E2-25y%2B144%3D0%5C%5C%5C%5CD%3D625-576%3D49%5C%5C%5C%5Cy_%7B1.2%7D%3D+%5Cfrac%7B25+%5Cpm+7%7D%7B2%7D%3B+y_1%3D16%3B+y_2%3D9%5C%5C%5C%5Cx_1%3D25-16%3D9%3B+x_2%3D25-9%3D16)
Ответ : числа 9 и 16
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Y = 1/<span>(6+x-x^2)^(1/3)
</span><span>6+x-x^2 </span>≠ 0
x^2 - x - 6 ≠ 0
x1 ≠ -2
x2 ≠ 3
D(y) = ( - ≈;-2) (-2;3) (3; + ≈)
Важно увидеть общий множитель и вынести его за скобки...