\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)          (điều kiện: \(x;y\ne0\); \(x\ne\pm2y\))

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

\(=\frac{2}{x+2y}+\frac{1}{x-2y}+\frac{4}{\left(x-2y\right)\left(x+2y\right)}\)

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

\(=\frac{3x-2y+4}{\left(x-2y\right)\left(x+2y\right)}\)

a) Xét tam giác ABD vuông tại A có:

AB²=BD²-AD² ( THEO định lý Pytago)

=> AB²=\(\left(4\sqrt{3}\right)^2-\left(2\sqrt{3}\right)^2=36\Rightarrow AB=6\left(cm\right)\)

OA=OB =1/2 BD=\(2\sqrt{3}\)

cHU VI tam giác AOB là: \(2\sqrt{3}+2\sqrt{3}+6=6+4\sqrt{3}\)

b)Tam giac AOD có OA=OD=AD=(\(2\sqrt{3}\)) nên tam giác AOD đều. => góc AOD=60⁰=> góc AOB=120⁰

góc ABO = góc BAO =(180⁰-120⁰):2=30⁰

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

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

3x^2-4x^2+2

=-x^2+2

= - ( x^2 -2 )

=\(-\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)\)

3x³ - 4x² + x

= x( 3x² - 4x + 1 )

= x( 3x² - 3x - x + 1 )

= x[ ( 3x² - 3x ) - ( x - 1 ) ]

= x[ 3x( x - 1 ) - ( x - 1 ) ]

= x( x - 1 )( 3x - 1 )

3x ³ - 4x ² + x

= x( 3x ² - 4x + 1 )

= x( 3x ² - 3x - x + 1 )

= x[ ( 3x ² - 3x ) - ( x - 1 ) ]

= x[ 3x( x - 1 ) - ( x - 1 ) ]

= x( x - 1 )( 3x - 1 )

Áp dụng BĐT \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) và \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\left(x,y,z>0\right)\) ta có :

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)

\(\ge\frac{\left(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\right)^2}{3}\ge\frac{\left(a+b+c+\frac{9}{a+b+c}\right)^2}{3}=\frac{\left(1+9\right)^2}{3}=\frac{100}{3}\)

Vậy \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{100}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)