<span>f(x)=x^3-2x^2+x+3
f`(x)=3x²-4x+1=3(x-1)(x-1/3)
3x²-4x+1=0
D=(-4)²-4*3*1=16-12=4=2²
x(1)=1
x(2)=1/3
f`(x)=0 при 3(x-1)(x-1/3)=0
+ - +
_________________1/3_____________1_____________
max min
x(max)=1/3 и х(min)=1 - стационарные точки</span>
Думаю так..............................
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A^2-9b^2=(a)^2-(3b)^2=(a-3b)(a+3b); x^2y^2-1=(xy)^2-1^2=(xy-1)(xy+1); 49x^2-121a^2=(7x)^2-(11a)^2=(7x-11a)(7x+11a); c^2d^2-m^2=(cd)^2-(m)^2=(cd-m)(cd+m).