Question

さく

25から全く意味がわかりません！！解説お願いします🙇🏻‍♂️💦一応付属の解説見ましたが全然わかりませんでした😢

25. 次の式を因数分解せよ.
(1) x2+2x+1-4y2
(0)
(2)x2-y2+2y-1
27. 次の式を最低次数の文字について整理して因数分解せよ.
(1) ax+x2+α-1
29. 次の式を因数分解せよ.
(1)x2+ (2y-1)x+ye-y
(2) ab+bc-b2-ac
(2)x2-(3y+1)x+2y2+3
31. 次の式を1つの文字で整理して因数分解せよ.
(1) 2(b-c) +62(c-a)+c2(a-b)
(2) 2(b+c) +62(c+α)+c2(a+b)+2abc
| 33. 次の式を適当な置き換えをして因数分解せよ.
(1)(x2+4x)2+7 (x²+4x) +12
(2) (2x-1)(x²-2x-4)+2
35. 次の式を因数分解せよ.
(1) x4-8x2+16
(3) x4 +3x2+4
37. 次の式を因数分解せよ.
(1)x3+1
(3)x-9x3+8
(2)x-9x2+8
(4) x4+x2+1
(2)x3-64
動画で解説
(3) (4)

22. (1) (5)=x+3x². (2y)+3+x-(2y)²+(2y)³
(2) (与式)
=x+6x2y+12xy²+8³
=(3a) +3. (3a)-2b+3-3a-(2b)²+(2b)³
=27a3+54a2b+36ab2+8b3
(3) (与式)
=(2x)³-3. (2x)2y+3+ (2x)·y²-y³
=8x³-12x²y+6xy2-y³
(4) (与式)
=(3a)-3-(3a)-4b+3.3a. (4b)²-(46)³
=27a3-108a2b+144ab2-64b³
23. (1) (5)=(x-2) (x²+x+2+22)=x³-23=x³-8
(2) (5)=(x+3)(x²-x+3+3²) = x²³+3³=x³+27
24. (1) (5)=(2x+y) | (2x)² - (2x)·y+y²
= (2x)³+y³= 8x³+y³
(2) (5)=(3a-2b)|(3a)+(3a) (2b)+(2b)²
= (3a)-(26)=27a³-863
25. (1) (5) = (x+1)² - (2y)²
=(x+1)+2y| | (x+1)-2y|
=(x+2y+1) (x-2y+1)
(2)与式)=(2-2y+1)
=x2(y-12
=(x+(y-1)x-(y-1)}
(x+y-1)(x-y+1)
26. (1) (5)=(x²-2xy+y²)-4z²
=(x-y)2-(2z)2
={(x-y)+2z}{(x-y)-2z}
=(x-y+2z)(x-y-2z)
(2) (与式) (4y2-12yz+9z2)
= x²-(2y-3z)²
e=x+(2y-32)} (x-(2y-3z)}
=(x+2y-32)(x-2y+32)
27. (1) (5)=(x+1)a+(x+1)(x-1)
=(x+1) (a+ (x-1)}
=(x+1)(a+x-1)
=(x+1)(x+a-1)
(2) (5)=(6-c)a-b(b-c)
=(a-b)(b-c)
28. (1) (5)=(a²-1)x+(a-1)
=(a-1)(a+1)x+(a−1)(
=(a-1){(a+1)x+1}
=(a-1)(ax+x+1)> (1)
(2) (5)=(a-b²)c+ab (a-b)=
=(a-b)(a+b)c+ab(a-b))
=(a-b)(a+b)c+abs()
=(a-b) (ab+be+ca)
29. (1) (5)=x+(2y-1)x+y(y-1)
=(x+y) (x+(y-1)]
=(x+y)(x+y-1)
(2) (5)=x²- (3y+1)x+(y+2)(2y-1)
={x-(y+2)|{x-(2y-1)}
=(x-y-2)(x-2y+1)
30. (1) (5)=2x²+(y-1)x-(y²-5y+6)
(参考)
X
=2x²+(y-1)x-(y-2)(y-3)
=x+(y-2) (2x-(y-3)}
=(x+y-2) (2x-y+3)
(y-2)-2y-4
2(y-3)-y+3
y-1
(2) (t) = 2x²+(5y-1)x+2(y²+2y-3)
=2x²+(5y-1)x+2(y+3)(y-1)
(参考)
(x+2(y-1) (2x+(y+3)}
= (x+2y-2) (2x+y+3)
2 (y-1)+4y-4
(y+3) y+3
5y-1
31. (1) (5)=(b-c) a²+b²c-b2a+ca²-bc²
(2+
=(b-c) a²- (62-c²) a+bc (b-c)
=(b-c) a²-(b-c)(b+c) a+bc (b-c)
=(b-c) {a²- (b+c) a+bc\
=(b-c)(a-b)(a–c)
=-(a-b)(b-c)(c-a)
(2) (5)=(b+c) a²+b²c+b²a+c²a+bc²+2abc
=(b+c) a²+(b2+2bc+c²) a+bc (b+c)
=(b+c) a²+ (b+c)2a+bc (b+c)
=(b+c) {a²+(b+c) a+bc}
=(b+c)(a+b)(a+c)
=(a+b)(b+c)(c+a)
32. (1) (5)=bc (b-c)-c2a-ca²+ab+ab²
=(b-c) a²+(62-c²) a+bc (b-c)
=(b-c) a²+(b+c) (b-c) a+bc (b-c)
(+= (b-c) {a²+(b+c) a+bc\
a)=(b-c)(a+b) (a+c)
=(a+b)(b-c)(c+a)
(2) (4) a² (b-c)+b2c+ab²+c²a-bc²-abc
=(b-c) a²+ (b2-bc+c²) a+bc (b-c)
=(a+(b-c)} {(b-c) a+bc)
=(a+b-c)(ab+bc-ca)
X (bc) - b²-2bc+c² (8)
(参考)
(4-1
(b-c) bc →
bc
b²-bc+c²
ート