Question

あ

高校からの課題で躓くところがありました。因数分解の問題です。良ければ教えてください全くわかりません。😭😭

35. (1) x²+x=y(y+1)
(3) x²-(y+3)x−(2y+1)(y−2)
(2)* x²-2xy +(y+1)(y−1)
□(4) x²+2xy+y²-5x-5y+6
教p.19 応用例題6

-25x2 に
35. (1) x²+x=y(y+1)
=(x−y)(x+y+1)
(2) x²−2xy+(y+1)(y−1),
{x-(y+1)}{x-(y-1)}
=(x-y-1)(x-y+1)
=
(3) x²-(y+3)x-(2y+1)(y-2)
{x-(2y+1)}{x+(y−2)}
=(x-2y-1)(x+y−2)
=
(4) x²+2xy+y²−5x−5y+6
=x²+(2y-5)x+y²-5y+6
=x²+(2y-5)x+(y-2)(y-3)
=(x+y-2)(x+y-3)
4
y →-y
X y+1→ y+1
1-y(y+1)
1
1-(y+1)→ -y-1
X=
1-(y-1)→-y+1
1 (y+1)(y-1) -2y
1-(2y+1)
y-2
y-2
1 −(2y+1)(y-2) -y-3
lyy-2
1y-3
→-2y-1
y-2
y-3
1 (y-2)(y-3) 2y-5