Có:\(x^4+64y^4\)

\(=\left(x^4+16x^2y^2+64y^4\right)-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+4xy+8y^2\right)\left(x^2-4xy+8y^2\right)\)

Linz

= 64y⁴ + 32xy³ + 8y²x²  - 32xy³  -16x²y² -  4x³y + 8x²y² +4x³y +x⁴

= 8y² ( 8y² + 4xy + x² ) - 4xy ( 8y² + 4xy + x² ) + x²  ( 8y² + 4xy + x² )

= ( 8y² - 4xy + x² ) ( 8y² + 4xy + x² )

\(A=\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}\)

\(=\frac{x^3}{xyz}+\frac{y^3}{xyz}+\frac{z^3}{xyz}\)

\(=\frac{x^3+y^3+y^3}{xyz}\)

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)-3xyz}{xyz}\)( HĐT x³ + y³ + z³ - 3xyz chắc biết rồi ha )

\(=\frac{-3xyz}{xyz}=-3\)( do x+y+z = 0 )

à nhầm dấu tí nhé ._.

\(A=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz}{xyz}=3\)

A = x² - 5x + 7 

= ( x² - 5x + 25/4 ) + 3/4

= ( x - 5/2 )² + 3/4 ≥ 3/4 ∀ x 

Dấu "=" xảy ra khi x = 5/2

=> MinA = 3/4, đạt được khi x = 5/2

a, Điều kiện xác định của A là : \(\hept{\begin{cases}3x-6\ne0\\x^2-2x\ne0\end{cases}\Rightarrow\hept{\begin{cases}3\left(x-2\right)\ne0\\x\left(x-2\right)\ne0\end{cases}\Rightarrow x\ne}0;2}\)

Vậy \(x\ne0;2\)

b, \(A=\frac{x}{3x-6}+\frac{2x+3}{x^2-2x}\)

\(=\frac{x}{3\left(x-2\right)}+\frac{2x+3}{x\left(x-2\right)}=\frac{x^2}{3x\left(x-2\right)}+\frac{6x+9}{3x\left(x-2\right)}=\frac{x^2+6x+9}{3x\left(x-2\right)}\)

c, Ta có : A = 0 hay 

\(\frac{x^2+6x+9}{3x\left(x-2\right)}=0\Leftrightarrow\left(x+3\right)^2=0\Leftrightarrow x=-3\)

Vậy x = -3 thì A = 0