Mathematics
Senior High
問5の(1)(2)の等式の照明の仕方を教えて欲しいです🙇♀️
a(a+b+c)=0
よって
(左辺) (右辺) = 0
-
したがって 2a²+bc = (b-a)(c-a)
問5 次の等式を証明せよ。
(1) x+y=1のとき
(2) a+b+c=0 のとき
x-x=y-y
p.62 Train
$65+ b = (58)
a+b2+c2 = -2 (ab+bc+ca)
a² b²+c²
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