Ta có M= \(x.y.x-x.y.y-5x+5y-12\)

=> \(M=5x-5y-5x+5y-12\)=-12

Vậy M=-12

\(a,=5\left(x-y\right)+a\left(x-y\right)=\left(5+a\right)\left(x-y\right)\\ b,=a\left(x+y\right)+b\left(x+y\right)=\left(a+b\right)\left(x+y\right)\\ c,=x\left(x+1\right)+a\left(x+1\right)=\left(x+a\right)\left(x+1\right)\\ d,Sửa:x^2y+xy^2-3x-3y=xy\left(x+y\right)-3\left(x+y\right)=\left(xy-3\right)\left(x+y\right)\\ e,=xy\left(x+1\right)-\left(x+1\right)=\left(xy-1\right)\left(x+1\right)\\ f,=x^2-4=\left(x-2\right)\left(x+2\right)\\ g,=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\\ h,=\left(x+5\right)^2-y^2=\left(x-y+5\right)\left(x+y+5\right)\\ i,=\left(x-4\right)^2-24y^2=\left(x-2\sqrt{6}y-4\right)\left(x+2\sqrt{6}y+4\right)\)

Bạn tham khảo nha:

M=x²y-xy²-5x+5y-12

M=xy(x-y)-(5x-5y)-12

M=5(x-y)-5(x-y)-12

M=0-12

M=-12

Chúc bạn học tốt nha!

Ta có: \(M=x^2y-xy^2-5x+5y-12\)

\(\Rightarrow M=x\times x\times y-x\times y\times y-5x+5y-12\)

\(\Rightarrow M=x\times5-5\times y-5x+5y-12\)

\(\Rightarrow M=\left(x\times5-5x\right)-\left(5\times y-5y\right)-12\)

\(\Rightarrow M=-12\)

Vậy khi \(xy=5\) thì \(M=-12\).

thì mik cứ để x=1; y=5 thôi. Thay vào tính => M=-12 chắc chắn

a/ \(P+Q=\left(x^2y+x^3-xy^2+3\right)+\left(x^3+xy^2-xy-6\right)\)

\(=x^2y+x^3-xy^2+3+x^3+xy^2-xy-6\)

\(=\left(x^3+x^3\right)+\left(xy^2-xy^2\right)+\left(3-6\right)+x^2y-xy\)

\(=2x^3+x^2y-xy-3\)

b/ \(M+N=\left(x^2y+0,5xy^3-7,5x^3y^2+x^3\right)+\)

\(\left(3xy^3-x^2y+5,5x^3y^2\right)\)

\(=x^2y+0,5xy^3-7,5x^3y^2+x^3+3xy^3-x^2y+5,5x^3y^2\)

\(=\left(x^2y-x^2y\right)+\left(0,5xy^3+3xy^3\right)+\left(5,5x^3y^2-7,5x^3y^2\right)+x^3\)

\(=3,5xy^3-2x^3y^2+x^3\)