Phân tích thành nhân tử à?
\(x^2-2xy+y^2+3x-3y-10\\ \\ =\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\\ =\left(x-y\right)^2+3\left(x-y\right)-10\\ =\left(x-y+3\right)\left(x-y\right)-10\left(1\right)\)
Đặt \(x-y=a\) \(\left(\text{*}\right)\)
Thay \(\left(\text{*}\right)\) vào \(\left(1\right)\)
\(\text{Ta được: }\left(1\right)=a\left(a+3\right)-10\\ \\ =a^2+3a-10\\ \\ =a^2+5a-2a-10\\ =\left(a^2+5a\right)-\left(2a+10\right)\\ \\ =a\left(a+5\right)-2\left(a+5\right)\\ \\ =\left(a-2\right)\left(a+5\right)\text{ }\text{ }\text{ }\left(2\right)\)
Thay \(\left(\text{*}\right)\) vào \(\left(2\right)\)
\(\text{Ta lại được: }\left(2\right)=\left(x-y-2\right)\left(x-y+5\right)\)
= (x-y)2 +3(x-y)-10
=(x-y).(x-y+3)-10
=(x-y)2+3(x-y)-10
thay x-y =a Ta có
a2-3a-10
=a2+2a-5a-10
=a(a+2)-5(a+2)
=(a-5)(a+2)
thay a ta có: (x-y-5)(x-y+2)
