Bài 4:
a:Sửa đề: \(A=\frac{a^2+8a+16}{a^2-2a-24}\)
\(=\frac{\left(a+4\right)^2}{a^2-6a+4a-24}\)
\(=\frac{\left(a+4\right)^2}{a\left(a-6\right)+4\left(a-6\right)}=\frac{\left(a+4\right)^2}{\left(a-6\right)\left(a+4\right)}=\frac{a+4}{a-6}\)
b: \(B=\frac{x-3}{x+3}+\frac{4x-24}{x^2-9}+\frac{4}{3-x}\)
\(=\frac{\left(x-3\right)^2+4x-24-4\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2-6x+9+4x-24-4x-12}{\left(x-3\right)\left(x+3\right)}=\frac{x^2-6x-27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x-9\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x-9}{x-3}\)
BÀi 6:
\(4x^2+2y^2+4xy-8y-6x+2020\)
\(=4x^2+4xy+y^2-6x-3y+y^2-5y+2020\)
\(=\left(2x+y\right)^2-3\left(2x+y\right)+\frac94+y^2-5y+\frac{25}{4}+2011,5\)
\(=\left(2x+y-\frac32\right)^2+\left(y-\frac52\right)^2+2011,5\ge2011,5\forall x,y\)
Dấu '=' xảy ra khi y-5/2=0 và 2x+y-3/2=0
=>y=5/2; 2x=-y+3/2=-5/2+3/2=-1
=>y=5/2; x=-1/2
