Question

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どうやって解くか途中式教えてください

32 次の式を因数分解せよ。
*(1) abx²-(α² +62)x+ab
(2) abx²+(a²-b²)xy-aby²

=)
3 → 12
-9-9
-27
3
2810b
-36-36
76
3)
-36)
= (a +2b)(a − 2b)(2a+3b)(2a-3b)
-16
-9-9
=
-25
36
4
30 (1) 与式=2629a²-462c²)
= 26²{(3a)²-(2bc)²}
= 26²(3a+2bc)(3a-2bc)
(2) 与式=
= (36a² — (b − c)²}={(6a)²-(b-c)²)
= {6a+ (b − c)}{6a — (b − c)}
= (6a+ b − c)(6a − b + c)
(3) 与式=2(x2+14xy-72y)
= 2(x²+(-4y+18y)x+(-4y). 18y)
= 2(x-4y)(x+18y)
(4) 与式=(x+a)(x-b)
(5) 5x=3x² +11xy+10y²-4y²
= 3x² +11xy+6y² = (x+3y)(3x+2y)
1
3y9y
3
2y2y
3
6y²
11y
31 (1) Git=(2x)3+13
= (2x+1){(2x)² - 2x·1+1²)
= (2x+1)(4x²-2x+1)
(2) =(4a)³-3³= (4a-3){(4a)² + 4a.3+3²)
= (4a-3)(16a²+12a+9)
(3) 与式=(3x)+(5y)3
= (3x+5y){(3x)²-3x-5y+(5y)²}
= (3x+5y) (9x²-15xy+25y²)
32 (1) 5 (ax-b)(bx-a)
→ b →
-62
X〓
- a
- a²
ab ab
-(a²+ b²)
(2) 5=(ax-by)(bx+ay)
x
-by-b²y
ay-
a'y
ab - aby²
(a²-b²) y
33 (1) 与式={(x2x-2}{(x2−x) - 12}
3²+ = (x²-x-2)(x² − x −12)
dr = (x+1)(x-2)(x+3)(x-4)
-66²
x-3)-y2
-2)²
-(2y-1)²
コリー (32) 2
2_16)
(3) = 4(a ²)²-25a²b²+36(6²)²
= (a²-46³) (4a²-96²)
(2) =(x²+2x){(x²+2x)-2)-3
= (x²+2x)²-2(x²+2x)-3
=((x²+2x)+1}{(x²+2x)-3)
= (x²+2x+1)(x²+2x-3)
=(x + 1)²(x-1)(x+3)
34 (1) 与式=x(y-1)-y+1
= x(y-1)-(y-1)=(x-1)(y-1)
別解 与式=(x-1)y-x+1
=(x-1)y-(x-1)=(x-1)(y-1) 0
= a(b-d)+bc-cd
(2)
= a(b-d) + c(b-d) = (a + c)(b-d)
参考 b c d のいずれで整理しても因数分解する
ことができる。
(3) 与式= (3x-15)y- (x2-25)
= 3(x-5)y-(x+5)(x-5)
=(x-5){3y-(x+5)}
=(x-5) (-x+3y-5)
(4) =(a²-1)b+(a²-1)=(a²-1)(b +1)
= (a + 1)(a-1)(b +1)
(5) 5=(2a +2b)c+(a²+2ab +6²)
(6) 与式=(2x-4)y+(2x2-3x−2)
=2(x−2)y+(x−2)(2x+1)
=(x-2){2y+ (2x+1)}
=(x-2)(2x+2y+1)
35 (1) = {x+(y+4)}{x+(2y-3)}
= (x+y+4)(x+2y-3)
X₂3+4 → y+4
2y-32y-3
3y+1
(2) 5t=x²+(3y-6)x+(2y²-11y+5)
=x²+(3y-6)x+(y-5X2y-1)
= {x+(y-5)}{x+(2y-1)}
= (x+y-5)(x+2y-1)
X₂
y-5→y-5
\2y-1→2y-1
3y-6
(3) 与式=x2-(2y+1)x + y^+ y - 2
=x²-(2y+1)x+(y-1)(y+2)
=(x-(y-1)}{x-(y+2)}
=(x-y+1)(x-y-2
1
-y+1
1 -(y+2) → -y-2
-2y-1
= 2(a+b)c+(a + b)² = (a+b){2c +(a+b)}
= (a + b)(a+b+2c)
別解 与式=(x-y)2
=(x-y
=(x-y
(4) 与式=2x2+ (51
=2x2+ (5
= (x+(2y
= (x+2y
与式=2x2+(
= 2x² +
F(x) = (x+(
= (x+3
2/
(6) 与式=2x2.
= 2x²
= (x+
=(x-
1
12
36 (1) 与式
=(b-c)a²
=(b-c)a-
=(b-c){a
=(b-c)(c
=-(a-l
注意 (b-c
a → b→
-(c-a)
(5)
(2) 与式
= {(a+b
= {a² +0
= (b + c
= (b + c
= {a+(
= (a +
1 +
b+