21. Phân tích A thành \(A=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(a^2+b^2+c^2+ab+bc+ac\right)\). Từ đó dễ dàng chứng minh.

23. \(9y\left(y-x\right)=4x^2\Leftrightarrow9y^2-9xy=4x^2\Leftrightarrow4x^2+9xy-9y^2=0\)

Chia cả hai vế của đẳng thức trên với \(y^2>0\)được : 

\(4\left(\frac{x}{y}\right)^2+\frac{9x}{y}-9=0\). Đặt \(t=\frac{x}{y},t>0\)(Vì x,y dương)

\(\Rightarrow4^2+9t-9=0\Leftrightarrow\orbr{\begin{cases}t=\frac{3}{4}\left(\text{nhận}\right)\\t=-3\left(\text{loại}\right)\end{cases}}\)

Vậy \(\frac{x}{y}=\frac{3}{4}\Rightarrow y=\frac{4x}{3}\)thay vào biểu thức được :

\(\frac{x-y}{x+y}=\frac{x-\left(\frac{4x}{3}\right)}{x+\left(\frac{4x}{3}\right)}=-\frac{1}{7}\)

Bài 1:

\(\frac{2}{x^2+2y^2+3}=\frac{2}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}\le\frac{2}{2xy+2y+2}=\frac{1}{xy+y+1}\)

Bài 2:

\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{52}{2x.3y}\ge\frac{16}{4x^2+9y^2+12xy}+\frac{52.4}{\left(2x+3y\right)^2}\)

\(A\ge\frac{16}{\left(2x+3y\right)^2}+\frac{208}{\left(2x+3y\right)^2}=\frac{224}{\left(2x+3y\right)^2}\ge\frac{224}{4}=56\)

\(A_{min}=56\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)

a) \(P=\dfrac{\left(x^2+2xy+9y^2\right)-\left(x+3y-2\sqrt{xy}\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)

\(=\dfrac{\left(x^2+6xy+9y^2\right)-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)

\(=\dfrac{\left(x+3y\right)^2-\left(x+3y\right)2\sqrt{xy}}{x+3y-2\sqrt{xy}}\)

\(=\dfrac{\left(x+3y\right)\left(x+3y-2\sqrt{xy}\right)}{x+3y-2\sqrt{xy}}\)

\(P=x+3y\)

b) \(\dfrac{P}{\sqrt{xy}+y}=\dfrac{x+3y}{\sqrt{xy}+y}=\dfrac{\left(x+3y\right):y}{\left(\sqrt{xy}+y\right):y}=\dfrac{\dfrac{x}{y}+3}{\sqrt{\dfrac{x}{y}}+1}\)

Đặt \(t=\sqrt{\dfrac{x}{y}}>0\) và \(\dfrac{P}{\sqrt{xy}+y}=Q\) thì \(Q=\dfrac{t^2+3}{t+1}=\dfrac{\left(t-1\right)^2+2\left(t+1\right)}{t+1}=2+\dfrac{\left(t-1\right)^2}{t+1}\ge2\)

\(Q_{min}=2\Leftrightarrow t=1\Leftrightarrow x=y\)

b) 

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

\(B=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}=\sqrt{x}+\sqrt{y}\)

\(A=\left(\dfrac{1}{\sqrt{x}+\sqrt{y}}+\dfrac{1}{\sqrt{y}-\sqrt{x}}\right):\dfrac{2\sqrt{xy}}{x-y}\)

\(=\dfrac{\sqrt{x}-\sqrt{y}-\sqrt{x}-\sqrt{y}}{x-y}:\dfrac{2\sqrt{xy}}{x-y}=\dfrac{-2\sqrt{y}}{2\sqrt{xy}}=\dfrac{-1}{\sqrt{x}}=\dfrac{-\sqrt{x}}{x}\)

b, Ta có \(A=\dfrac{-1}{\sqrt{x}}=1\Leftrightarrow\sqrt{x}=-1\left(voli\right)\)

Vậy pt vô nghiệm 

B1 : 

Áp dụng bđt cosi ta có : a^2/b+c + b+c/4 >= \(2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}\) = 2. a/2 = a

Tương tự b^2/c+a + c+a/4 >= b

c^2/a+b + a+b/4 >= c

=> VT + a+b+c/2 >= a+b+c

=> VT >= a+b+c/2 = VP 

=> ĐPCM

Dấu "=" xảy ra <=> a=b=c > 0

k mk nha

\(A=\frac{4\left(x+y+\sqrt{xy}\right)}{x+y+2\sqrt{xy}}=\frac{3\left(x+y+2\sqrt{xy}\right)+\left(x+y-2\sqrt{xy}\right)}{\left(x+y+2\sqrt{xy}\right)}=\frac{3\left(\sqrt{x}+\sqrt{y}\right)^2+\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2}+3\ge3\)

=> \(A\ge3\)

Vậy Min A = 3 khi x=y