\(1,3x\left(2x^2+1\right)=6x^3+3x\left(B\right)\\ 2,\left(x^4-2x^3+4x^2\right):2x^2=\dfrac{1}{2}x^2-x+2\left(B\right)\\ 3,\left(3x+2y\right)\left(2x+3y\right)=6x^2+9xy+4xy+6y^2=6x^2+13xy+6y^2\left(D\right)\\ 4,\left(x-2\right)^3=x^3-6x^2+12x-8\left(C\right)\\ 5,P=x^2-2x+2=\left(x-1\right)^2+1=\left(1-1\right)^2+1=0+1=1\left(D\right)\\ 6,\widehat{D}=360^0-\widehat{A}-\widehat{B}-\widehat{C}=360^0-120^0-80^0-100^0=60^0\left(D\right)\\ 7,B\)
\(8,\)
Chứng minh được EF,EI,KF lần lượt là đường trung bình hình thang ABCD; tam giác ABD; tam giác ABC
\(\Rightarrow EF=\dfrac{AB+DC}{2}=5;EI+KF=\dfrac{AB}{2}+\dfrac{AB}{2}=AB=3\\ \Rightarrow IK=EF-EI-KF=5-3=2\left(C\right)\)
