Question

ゆうな

途中式もお願いします！

因数分解

Answer 1

✨ ベストアンサー ✨

ぽーん

(1)3ab-a+3b-1 (2)8xy-2x-4yz+z
=(3b-1)a+(3b-1) =2x(4y-1)-(4y-1)z
=(3b-1)(a+1) =(4y-1)(2x-z)

(3)a²-2ab+2b-1 (4)m²+mn-4m-n+3
=(a²-1)-(2ab-2b) =(m²-4m+3)+(mn-n)
=(a+1)(a-1)-2b(a-1) =(m-3)(m-1)+(m-1)n
=(a-1){(a+1)-2b} =(m-1){(m-3)+n}
=(a-1)(a-2b+1) =(m-1)(m+n-3)

(5)x²y-x²z+y²z-xy² (6)xyz+x²y-xy²-x+y-z
=(x²y-xy²)-(x²z-y²z) =(x²y-xy²)-(x-y)+(xyz-z)
=(x-y)xy-(x²-y²)z =(x-y)xy-(x-y)+(xy-1)z
=(x-y)xy-(x+y)(x-y)z =(x-y)(xy-1)+(xy-1)z
=(x-y){xy-(x+y)z} =(x-y)(xy-1)(1+z)
=(x-y)(xy-yz-zx)

(7)2x²-3xy-2y²+5x-5y+3
=2x²+(-3xy+5x)-(2y²+5y-3)
=2x²+(-3y+5)x-(y+3)(2y-1)
={x-(2y-1)}{2x+(y+3)}
=(x-2y+1)(2x+y+3)

(8)2x²-3xy+y²+5x-2y-3
=2x²+(-3xy+5x)+(y²-2y-3)
=2x²+(-y+5)x+(y+1)(y-3)
={x+(y-3)}{2x+(y+1)}
=(x+y-3)(2x+y+1)

(9)3x²-5xy-2y²+x+5y-2
=3x²+(-5xy+x)-(2y²-5y+2)
=3x²+(-5y+1)x-(y-2)(2y-1)
={x-(2y-1)}{3x+(y-2)}
=(x-2y+1)(3x+y-2)

(10)6x²-xy-y²+13x-4y+5
=6x²+(-xy+13x)-(y²+4y-5)
=6x²+(-y+13)x-(y+5)(y-1)
={2x-(y-1)}{3x+(y+5)}
=(2x-y+1)(3x+y+5)

ゆうな

ありがとうございます😊