<span>x</span>² <span>+ 3x</span>⁴ = х²(1 + 3х²)<span>;
3a</span>² <span>- 27 = 3(а</span>² - 9) = 3(а - 3)(а + 3)<span>;
2x + 4 + x</span>² <span>+ 2x = 2(х + 2) + х(х + 2) = (х + 2)( 2 + х) = (х + 2)</span>²<span>;
Решите уравнение:
X</span>² <span>- 3x = 0
х(х - 3) = 0
х = 0 или х - 3 = 0
х = 3
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1)6(х+у)
4(5а-b)
x(c+d)
2)(x+a)(x+a)
(2a-5b)(2a-5b)
(3p-8)(3p-8)
3)7xy(2x-y+3xy)
20a{в четвертой} b³(2+4b-3ab²)
4)2m(a-b) - 3n(a-b) = (a-b)(2m-3n)
(a-14)(a-14-7) = (a-14)(a-21)
x^2 - (x^2 - 1^2) = x^2 - x^2 + 1 = 1
Решение во вложенииииииииииииииииииииииииииииииии