a. 

b.

c. 

a.\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[\left(x+y\right)^3+z^3\right]-a^3-b^3-c^3\)

\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)

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

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

b.\(x^4+2010x^2+2009x+2010\)

\(=\left(x^4-x\right)+\left(2010x^2+2010x+2010\right)\)

=\(x\left(x-1\right)\left(x^2+x+1\right)+2010\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^2-x+2010\right)\)

1. Ta có:

(x-y)² =5²=25.

\(\Rightarrow\)x²-2xy+y²=25 \(\Rightarrow\)-2xy=25-55=-30( vì x²+y²=55).

\(\Rightarrow\)xy=15.

Ta có x²+2xy+y²=55+15=70

\(\Rightarrow\)(x+y)²=70

S_ABD=\(\dfrac{1}{2}\)AD.AB=\(\dfrac{1}{2}\)SABCD.

S_AFB=\(\dfrac{1}{2}\)S_ABD (vì chung chiều cao hạ từ A và FB=\(\dfrac{1}{2}\)BD).

S_AEF=\(\dfrac{1}{2}\)S_AFB (vì chung chiều cao hạ từ F và EA=\(\dfrac{1}{2}\)AB).

\(\Rightarrow\)S_AEF=\(\dfrac{1}{2}\).\(\dfrac{1}{2}\).\(\dfrac{1}{2}\)S_ABCD=\(\dfrac{1}{8}\)S_ABCD.

mik cx vậy.Violympic làm ăn thật!!!!!!!!!!!!!!

program dientich;

uses crt;

var a,b:integer;

S:longint;

begin

clrcsr;

write('nhap a='); readln(a);

write('nhap b='); readln(b);

S:=(a*b)/2;

write('dien tich hinh thoi=',S);

readln;

end.