\(\left\{{}\begin{matrix}a,x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\\b,x^2-2xy+y^2-25=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\\c,7x^2-14x+7=7\left(x^2-2x+1\right)=7\left(x-1\right)^2\\d,x^2+4x+4-y^2=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}e,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\f,x^2+2xy+y^2-9=\left(x+y\right)^2-9=\left(x+y-3\right)\left(x+y+3\right)\\g,x^2y-2xy+y=y\left(x^2-2x+1\right)=y\left(x-1\right)^2\\h,x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}k,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\m,5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\end{matrix}\right.\)Câu trả lời: \(\left\{{}\begin{matrix}a,x^4+2x^3+x^2=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\\b,x^2-2xy+y^2-25=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\\c,7x^2-14x+7=7\left(x^2-2x+1\right)=7\left(x-1\right)^2\\d,x^2+4x+4-y^2=\left(x+2\right)^2-y^2=\left(x+2-y\right)\left(x+2+y\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}e,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\f,x^2+2xy+y^2-9=\left(x+y\right)^2-9=\left(x+y-3\right)\left(x+y+3\right)\\g,x^2y-2xy+y=y\left(x^2-2x+1\right)=y\left(x-1\right)^2\\h,x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}k,x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\m,5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\end{matrix}\right.\)