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`

$=[(y-4)(y^2-10y+25-y+5+1)]^2$

$=[(y-4)(y^2-11y+31)]^2$

$=(y-4)^2.(y^2-11y+31)^2$

Đáp án:

$(5−y)^6−2(125−75y+15y^2−y^3)+1$

$=(5−y)^6−2( 5^3 - 3.5^2.y + 3.5.y^2 - y^3) + 1$

$= (5 - y)^6 - 2.(5 - y)^3 + 1 $

$=[(5−y)^3−1]^2$

Đặt t = 5 - y

=> $=[(5−y)^3−1]^2$ 

$= ( t^3 - 1)^2$

$ = ( t^3 - 1^3)^2$

$ = [( t - 1)(t^2 - t + 1)]^2$

$ = (t - 1)^2.(t^2 - t + 1)^2$

Thay vào ta được

$ = ( 5 - y - 1)^2[(5 - y)^2 - (5 - y) + 1]^2$

$ = ( 4 - y)^2.(y^2 - 11y + 31)^2$

Giải thích các bước giải: