Đáp án: $(y-4)^2.(y^2-11y+31)^2$
Giải thích các bước giải:
Cách 1:
$(5-y)^6-2(125-75y+15y^2-y^3)+1$
$=(y-5)^6+2(y^3-15y^2+75y-125)+1$
$=(y-5)^6+2(y^3-3.y^2.5+3.y.5^2-5^3)+1$
$=(y-5)^6+2(y-5)^3+1$
$=[(y-5)^3+1]^2$
`={(y-5+1)[(y-5)^2-(y-5)+1]}^2`
$=[(y-4)(y^2-10y+25-y+5+1)]^2$
$=[(y-4)(y^2-11y+31)]^2$
$=(y-4)^2.(y^2-11y+31)^2$
Cách 2:
$(5-y)^6-2(125-75y+15y^2-y^3)+1$
$=(y-5)^6+2(y^3-15y^2+75y-125)+1$
$=(y-5)^6+2(y^3-3.y^2.5+3.y.5^2-5^3)+1$
$=(y-5)^6+2(y-5)^3+1$
$=[(y-5)^3+1]^2$
Đặt $y-5=x$
Như vậy:
$[(y-5)^3+1]^2$
$=(t^3-1)^2$
$=[(t-1)(t^2+t+1)]^2$
$=(t-1)^2.(t^2-t+1)^2$
`={(y-5+1)[(y-5)^2-(y-5)+1]}^2`
$=[(y-4)(y^2-10y+25-y+5+1)]^2$
$=[(y-4)(y^2-11y+31)]^2$
$=(y-4)^2.(y^2-11y+31)^2$
Cách 3:
Đặt $(5-y)^3=x=125-75y+15y^2-y^3)+1$
Như vậy:
$(5-y)^6-2(125-75y+15y^2-y^3)+1$
$=x^2-2x+1=(x-1)^2$
$=[(5-y)^3-1]^2$
$=[(y-5)^3+1]^2$
`={(y-5+1)[(y-5)^2-(y-5)+1]}^2`
$=[(y-4)(y^2-10y+25-y+5+1)]^2$
$=[(y-4)(y^2-11y+31)]^2$
$=(y-4)^2.(y^2-11y+31)^2$
