Question

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33の(2)(4)(5)がよく分からないです😭😭💦
どなたか教えてください🙇🏻‍♀️🙇🏻‍♀️😭

=(x-2)-(x-2y+2))
=-(x-2)(x-2y+2)
(4) bについて整理すると
(5)=(a²-1)b+(a-1) r
= (a +1Xa-1)b+(a−1)
= (a-1)(a +1)b+1)
=(a-1Xab+b+1)
(5) cについて整理すると
(5x)=(b-a)c+(a²-2ab+6²)
1 = (a - b)(a-b-c)
(6) z について整理すると
(与式)=(-4x2+y2z + (4x²y-y3)
= -(4x² - y²)2+(4x² - y²) y
= (4x² - y²)(-2+ y)
=(2x+y)(2x-y) (y-z)
33 (1) (t)={x+(2y-1)}{x+(3y+2)}
= (x+2y-1)(x+3y+2)
=-(a-b)c+(a-b)²
= (a−b){-c+(a−b)}
→2y-1
3y+2
1 (2y-1)(3y+2) 5y+1 =
(2) (5)=x²+(-a-5)x-(2a²-a-6)
1
ix
1
a-2
-(2a +3)
1 -(a-2)(2a+3)
(3) xについて整理すると
(x)=x²+(-3y-1)x+(2y²+5y-12)
=x²+(-3y-1)x+(y+4)(2y-3)
=(x-(y+4)}{x-(2y-3))}
=(x-y-4)(x-2y+3)
-(y+4)
-(2y-3)
(y+4)2y-3)
1
2
2y-1
3y+2
= x² + (-a-5)x-(a-2)X2a+3)
{x+(a-2)}{x-(2a+3)}
=
= (x+a-2)(x-2a-3)+
a-2
→ -2a-3
-a-5
1
(4) xについて整理すると
(与式)=2x2+(y-3)x-(y2-1)
--y-4
-2y+3
-3y-1
= 2x² + (y-3)x-(y+1Xy-1)
= {x+(y-1)}{2x-(y+1)}
=(x+y-1)(2x-y-1)
→
-(y+1)
-(y+1)(y-1)
y-1→
2y-2
-y-1
y-3
(5) xについて整理すると
(与式)=6x²+(-7y-6)x+2y^+5y-12)
=6x²+(-7y-6)x+(y+4)2y-3)
=(2x-(y+4)(3x-(2y-3))
=(2x-y-4x3x2y+3)
2
3X (9+4)
6
34 (1) (与式)
-(2y-3)
(y+4)(2x-3)
--3y-12
--4y+6
-79-6
=a²b+ab² + b²c+bc²+c²a+ca³ + 2abc
= (b+c)a²+(b²+2bc+c²a+b³c+bc²
= (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)
(2) (与式)
= (b + c)a² + (b²+3bc+c²)a+(b²c+bc²)
=
(b+c)a²+(b²+3bc+c²)a+bc(b+c)
= 1. (b+c)a²+{1·bc+(b + c)²}a +(b+c)-bc
= {a+(b + c)(b + c)a+bc}
= (a+b+c)(ab+bc+ca)
b+c
bc
bc(b + c)
otex
b+c
→ b²+2bc+c²
bc
b²+3bc+c²
35 (1) (与式)=α3-3・a²2+3 ・α・22-23
=a³-6a²+12a-8
(2) (与式)(3x)+3(3x) 2・1+3・3x・12+13
= 27x³+27x² +9x+1
(3) (与式)=(2x)+3 (2x)^2-3y+3.2x ・ ( 3y)2+(3y)
= 8x³ +36x²y+54xy² +27y³
(4) (t)=(4a)³-3-(4a)²-3b+3.4a (36)²-(3b)³
=64a³-144a²b+108ab²-27b³
(5) (与式)=(x+3)(x2-x ・3+32)
= x³ +3³ = x³+27
(6) (t)=(a-1Xa²+a-1+1²)
=a³-1³-a³-1
(7)
(5)=(2a+b){(2a)²-2a-b+b²)
= (2a)³ + b³=8a³ + b³
(8) (t) = (3x-5y){(3x)²+3x-5y+(5y)²)
= (3x)³ (5y)³=27x³-125y³
36 (1) (与式)=x-4°= (x-4)(x2+x 4 +4%)
=(x-4)(x²+4x+16)
(2) (t)=(2a)³ +3³ = (2a+3){(2a)²-2a-3+3²
= (2a+3)(4a²-6a +9)