a)\(\dfrac{x}{y}+\dfrac{y}{x}-2=\dfrac{x^2+y^2-2xy}{xy}=\dfrac{\left(x-y\right)^2}{xy}\)\(\ge0\)

Vậy \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

b) ta có: A=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)=\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\)

A\(\ge\)\(\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)-2\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+2\)

=\(\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2\ge0\)

c) Từ câu b suy ra:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)\ge0\)

Vì \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)(câu a)

Nên:

\(\left(\dfrac{x^4}{y^4}+\dfrac{y^4}{x^4}\right)-\left(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\right)+\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge2\)

\(a,VT=\dfrac{x^2+2xy+4-3x^2-3xy}{\left(x+y\right)\left(x+2y\right)}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x-2y\right)}=VP\\ b,VP=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}=VT\)

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)

b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)

c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

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

a.

Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)

Thế vào \(2x+y-z=81\)

\(\Rightarrow2.5k+3k-4k=81\)

\(\Rightarrow9k=81\)

\(\Rightarrow k=9\)

\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)

b.

Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)

Thế vào \(5x-y+3z=124\)

\(\Rightarrow5.3k-5k+3.2k=124\)

\(\Rightarrow16k=124\)

\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)

c.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Thế vào \(xyz=810\)

\(\Rightarrow2k.3k.5k=810\)

\(\Rightarrow k^3=27\)

\(\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)

d.

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)

Thế vào \(x^2y^2z^2=288^2\)

\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)

\(\Rightarrow\left(k^2\right)^3=64\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)

đúng rùi đó

Sai

a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)

\(=\dfrac{1}{6}\sqrt{6}\)

b: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)

\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)

⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)

⇒x=70;y=105;z=84

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)⇒\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{z^2}{25}=\dfrac{x^2-2y^2+z^2}{4-18+25}=\dfrac{44}{11}=4\)

⇒x=8;y=12;z=20

Nesbit:v dài

Nham ko phai Nesbit, Cauchy-Schwarz ra luon