Mathematics
Senior High
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別解のところの(a+b)(b+c)(c+a)=(3-a)(3-b)(3-c)は公式ですか?
(4) a+β+y=3から
a+8=3-r, B+r=3-a, r+a=3-ß
したがって
(a+B)(B+r)r+a)=(3-a)(3-B)(3-r)
① の両辺にx=3 を代入して
3³-3.32+2-3+4=(3-a)3-8)(3-7)
よって
(3-a)(3-B)(3-7)=10
すなわち (α+β(+r)(y+α) = 10
(a+B)(B+r) (r+a)
Seve
= (3-a)(3-B)(3-r) = {
=(9-38-3a+aß)(3-r)
= 27-9(a +8+r)+3(aß+Br+ra) - aßr
=27-9・3+3.2- (-4)=10
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公式は無いんですね!ありがとうございます🙇♀️