(5)
(与式)
=2c(b-a)-(a^2-2ab+b^2)
=2c(b-a)-(a-b)^2
=(a-b){-2c-(a-b)}
=(a-b)(-a+b-2c)
(6)
(与式)
=xy(x-y)-z(x^2-y^2)
=xy(x-y)-z(x+y)(x-y)
=(x-y)(xy-yz-zx)
至急!!(5)(6)を教えてください!
答えは(5)は(a-b)(-a+b-2c)
(6)は(x-y)(xy-yz-zx)です!
お願いします🙇🙇🙇
(5)
(与式)
=2c(b-a)-(a^2-2ab+b^2)
=2c(b-a)-(a-b)^2
=(a-b){-2c-(a-b)}
=(a-b)(-a+b-2c)
(6)
(与式)
=xy(x-y)-z(x^2-y^2)
=xy(x-y)-z(x+y)(x-y)
=(x-y)(xy-yz-zx)
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(5)の3行めからわかりません!
2乗はどこいったんですか??