a^3-6a^2b+12ab^2-8b^3=a^2(a-2b)-4a^2b+12ab^2-8b^3=a^2(a-2b)-2ab(a-2b)+4b^2(a-2b)=(a-2b)(a^2-4ab+4b^2)=(a-2b)(a-2b)^2=(a-2b)^3
В решении.
Объяснение:
а) Чтобы определить принадлежность точки графику, нужно известные значения х и у (координаты точки) подставить в уравнение. Если левая часть равна правой, то принадлежит, и наоборот.
у=√х
1) А(49; -7)
-7 = ± √ 49
-7 = -7, проходит.
2) В(0,09; 0,3)
0,3 = √0,09
0,3 = 0,3, проходит.
3) С(63; 3√7)
3√7 = √63
3√7 = √9*7
3√7 = 3√7, проходит.
б) х ∈ [0; 25]
y=√0 = 0;
y=√25 = 5;
При х ∈ [0; 25] у ∈ [0; 5].
в) у ∈ [9; 17]
у = √х
9=√х х=9² х=81;
17=√х х=17² х=289.
При х ∈ [81; 289] у ∈ [9; 17].
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Проверим, применив формулу