\(5x^2-5xy+y^2+\frac{4}{x^2}=0\Leftrightarrow4x^2-4xy+y^2+x^2+\frac{4}{x^2}-4+4=xy\)

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

Vì vế trái >= 4 với mọi x, y => vế phải >=4

Dấu = xảy ra \(\Leftrightarrow y=2x\)và \(x=\frac{2}{x}\)\(\Leftrightarrow x=\sqrt{2}\)và \(y=2\sqrt{2}\) hay \(x=-\sqrt{2}\)và \(y=-2\sqrt{2}\)

\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

                                             \(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

bài 3 min hay max ?

Đặt \(B=xy=2013-A\) thế vô cái cần tìm thì được

\(5x^2+\frac{y^2}{4}+\frac{1}{4x^2}=\frac{5}{2}\)

\(\Leftrightarrow x^2y^2+20x^4-10x^2+1=0\)

\(\Leftrightarrow20x^4-10x^2+1+B^2=0\)

\(\Leftrightarrow B^2=\frac{1}{4}-\left(\sqrt{20}x^2-\frac{\sqrt{5}}{2}\right)^2\le\frac{1}{4}\)

\(\Leftrightarrow-\frac{1}{2}\le B\le\frac{1}{2}\)

\(\Leftrightarrow-\frac{1}{2}\le2013-A\le\frac{1}{2}\)

\(\Leftrightarrow2012,3\le A\le2013,5\)

bạn chưa ghi gtnn , gtln xảy ra khi x=? và y=?

lon ok

Yes.

Dăm ba mấy dạng kiểu này

Cô si vài cái có gì khó khăn !

\(P=\frac{3yz}{x}+\frac{4xz}{y}+\frac{5xy}{z}\)

\(=\left(\frac{yz}{x}+\frac{zx}{y}\right)+2\left(\frac{yz}{x}+\frac{xy}{z}\right)+3\left(\frac{zx}{y}+\frac{xy}{z}\right)\)

\(\ge2z+3y+6x\)

\(=2\left(x+z\right)+4\left(x+y\right)\)

\(\ge4\sqrt{xz}+8\sqrt{xy}\)

\(=4\left(\sqrt{xz}+2\sqrt{xy}\right)\)

\(=4\)

Dấu "=" xảy ra tại \(x=y=z=\frac{1}{3}\)

`@` `\text {Ans}`

`\downarrow`

`x^2 + xy - 2x - 2y`

`= (x^2 - 2x) + (xy - 2y)`

`= x(x - 2) + y(x - 2)`

`= (x + y)(x - 2)`

____

`x^2 - xy - 6x + 6y`

`= (x^2 - 6x) - (xy - 6y)`

`= x(x - 6) - y(x - 6)`

`= (x - y)(x - 6)`

____

`5xy^2 - 5x + y^2 - 1`

`= (5xy^2 + y^2) - (5x + 1)`

`= y^2(5x + 1) - (5x + 1)`

`= (y^2 - 1)(5x + 1)`

`= (y - 1)(y + 1)(5x + 1)`

a: =(x^2+xy)-(2x+2y)

=x(x+y)-2(x+y)

=(x+y)(x-2)

b: =(x^2-xy)-(6x-6y)

=x(x-y)-6(x-y)

=(x-y)(x-6)

c: =5xy^2+y^2-5x-1

=y^2(5x+1)-(5x+1)

=(5x+1)(y^2-1)

=(5x+1)(y+1)(y-1)