xét ▲ABC có EB=EA;FA=FC≫EF la duờng trung binh 

≫EF//BC

≫tứ giác EFBC là hinh thang

ME//AC mà MB=MC ≫EB=EA

cmtt,FA=FC

\(\frac{4x^2-16}{x^2+2x}=\frac{A}{x}\)

\(\Rightarrow A.\left(x^2+2x\right)=\left(4x^2-16\right).x\)

\(\Rightarrow A=\frac{\left[\left(2x\right)^2-4^2\right].x}{x^2+2x}\)

\(A=\frac{\left(2x-4\right)\left(2x+4\right).x}{x\left(x+2\right)}\)

\(A=\frac{2.2.\left(x-2\right)\left(x+2\right).x}{x\left(x+2\right)}\)

\(A=4\left(x-2\right)\)\(\left(x\ne0;x+2\ne0\right)\)

\(A=4x-8\)

Ta có: \(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)

\(\Rightarrow a^3+b^3+c^3=-3ab.\left(-c\right)=3abc\)

Mặt khác: \(a+b+c=0\Rightarrow a^2=\left(-b-c\right)^2=\left(b+c\right)^2\)

\(\Rightarrow a^2-b^2-c^2=\left(b+c\right)^2-b^2-c^2=2bc\)

Tương tự ta có: \(b^2-a^2-c^2=2ca\)

\(c^2-a^2-b^2=2ab\)

\(\Rightarrow B=\frac{a^2}{2ab}+\frac{b^2}{2ca}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)

\(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-c\right)\left(a-b\right)\left(b-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=0\)

\(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{-a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{-ab+ac-bc+ab-ac+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=0\)

\(x^2-2x+\left(x-2\right)^2\)

\(=x^2-2x+x^2-4x+4\)

\(=2x^2-6x+4\)

\(=2.\left(x^2-3x+2\right)\)

\(=2.\left[\left(x^2-x\right)-\left(2x-2\right)\right]\)

\(=2.\left[x.\left(x-1\right)-2.\left(x-1\right)\right]\)

\(=2.\left(x-1\right)\left(x-2\right)\)

\(a,x^2-2x+\left(x-2\right)^2\)

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

\(=\left(x+x-2\right)\left(x-2\right)\)

\(b,x^2-6xy-16+9y^2\)

\(=\left(x^2-6xy+9y^2\right)-16\)

\(=\left(x+3y\right)^2-4^2\)

\(=\left(x+3y-4\right)\left(x+3y+4\right)\)