Trả lời:

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

b, \(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{y}+\sqrt{x}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+2\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(2\sqrt{y}\right)^2\)

\(=x-4y\)

Từ pt đã cho dễ dàng suy ra x,y>0

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}\cdot\sqrt{x}}=2\)

\(\frac{1}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{\frac{1}{\sqrt{y}}\cdot\sqrt{y}}=2\)

Cộng theo vế 2 BĐT trên ta có:

\(VT=\frac{1}{\sqrt{x}}+\sqrt{x}+\frac{1}{\sqrt{y}}+\sqrt{y}\ge4=VP\)

Khi \(x=y=1\)

mọi người giúp mình với mình cần gấp lắm

a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2\)

b,\(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)\)

\(=\left(2\sqrt{x}\right)^2-\left(\frac{2}{3}\right)^2\)

Trả lời:

a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)\)\(=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2=\frac{x^2}{y^2}-\frac{4}{9}\)

b, \(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)=\left(2\sqrt{x}\right)^2-\left(\frac{2}{3}\right)^2=4x-\frac{4}{9}\)

Bài 3: 

a: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x-4\right)\left(x+4\right)=21\)

\(\Leftrightarrow x^3-27-x\left(x^2-16\right)=21\)

\(\Leftrightarrow x^3-27-x^3+16x=21\)

=>16x=48

hay x=3

b: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=4\)

\(\Leftrightarrow x^3+8-x^3-2x=4\)

=>-2x=4-8=-4

hay x=2

\(=>P=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\sqrt{3}\)

CHÚC BẠN HỌC TỐT..........