\(\sqrt{x+y-4}+\sqrt{x-y+4}+\sqrt{-x+y+4}=\sqrt{x}+\sqrt{y}+2\)

ĐKXĐ:

 \(x+y-4\ge0\rightarrow x+y\ge4\rightarrow x+y\ge4\)

\(x-y+4\ge0\rightarrow x-y\ge-4\rightarrow x-y\ge-4\)

\(-x+y+4\ge0\rightarrow-x+y\ge-4\rightarrow x-y\le4\)

\(x\ge0\)

\(y\ge0\)

Với \(a;b\ge0\) ta có:

\(\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\)

\(\Leftrightarrow a-2\sqrt{ab}+b\ge0\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow2\left(a+b\right)\ge a+b+2\sqrt{ab}\)

\(\Leftrightarrow2\left(a+b\right)\ge\left(\sqrt{a}+\sqrt{b}\right)^2\)

\(\rightarrow\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}\)

Đẳng thức xảy ra khi \(a=b\)

Áp dụng bất đẳng thức trên, ta có:

\(\hept{\begin{cases}\sqrt{x+y-4}+\sqrt{x-y+4}\le\sqrt{2\left(x+y-4+x-y+4\right)}=2\sqrt{x}\\\sqrt{x+y-4}+\sqrt{-x+y+4}\le\sqrt{2[\left(x+y-4\right)+\left(-x+y+4\right)]}=2\sqrt{y}\\\sqrt{x-y+4}+\sqrt{-x+y+4}\le\sqrt{2[\left(x-y+4\right)+\left(-x+y+4\right)}=4\end{cases}}\)

\(\rightarrow2\sqrt{x+y-4}+2\sqrt{x-y+4}+2\sqrt{-x+y+4}\le2\sqrt{x}+2\sqrt{y}+4\)

\(\rightarrow\sqrt{x+y-4}+\sqrt{x-y+4}+\sqrt{-x+y+4}\le\sqrt{x}+\sqrt{y}+2\)

Đẳng thức xảy ra khi \(\hept{\begin{cases}x+y-4=x-y+4\\x+y-4=-x+y+4\\x-y+4=-x+y+4\end{cases}}\Leftrightarrow x=y=4\) (Thoả mãn)

\(x\ne1;x\ge0\)

\(A=x+2\) 

\(mincủaA=2khix=0\)

are you sad? 🥺🥺🥺

He wondered how he could solve that problem.

Điều kiện \(x+y\ge0\) và \(x\ge y\)

Xét phương trình thứ hai: \(\sqrt{\frac{x+y}{8}}-\sqrt{\frac{x-y}{12}}=3\)\(\Leftrightarrow\frac{1}{2}\sqrt{\frac{x+y}{2}}-\frac{1}{2}\sqrt{\frac{x-y}{3}}=3\)

\(\Leftrightarrow\sqrt{\frac{x+y}{2}}-\sqrt{\frac{x-y}{3}}=6\)

Như vậy hệ đã cho \(\Leftrightarrow\hept{\begin{cases}\sqrt{\frac{x+y}{2}}+\sqrt{\frac{x-y}{3}}=14\\\sqrt{\frac{x+y}{2}}-\sqrt{\frac{x-y}{3}}=6\end{cases}}\)(*)

Đặt \(\sqrt{\frac{x+y}{2}}=a\left(a\ge0\right)\)và \(\sqrt{\frac{x-y}{3}}=b\left(b\ge0\right)\), khi đó 

(*) \(\Leftrightarrow\hept{\begin{cases}a+b=14\\a-b=6\end{cases}}\Leftrightarrow\hept{\begin{cases}2a=20\\b=a-6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=10\\b=10-6=4\end{cases}}\)(nhận)

Vậy \(\sqrt{\frac{x+y}{2}}=10\)\(\Leftrightarrow\frac{x+y}{2}=100\)\(\Leftrightarrow x+y=200\)

và \(\sqrt{\frac{x-y}{3}}=4\)\(\Leftrightarrow\frac{x-y}{3}=16\)\(\Leftrightarrow x-y=48\)

Vậy hệ đã cho \(\Leftrightarrow\hept{\begin{cases}x+y=200\\x-y=48\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=248\\y=x-48\end{cases}}\Leftrightarrow\hept{\begin{cases}x=124\\y=124-48=76\end{cases}}\)(nhận)'

Vậy hệ phương trình đã cho có nghiệm duy nhất là \(\left(124;76\right)\)

Khi nào rảnh vào kênh H-EDITOR đăng kí   nha!!! Thanks!

\(f\left(x\right)=\dfrac{12\left(x^2+5,76\right)}{4\sqrt{x^2+3,24}.3\sqrt{x^2+10,24}}=\dfrac{12\left(x^2+5,76\right)}{\sqrt{16x^2+51,84}.\sqrt{9x^2+92,16}}\)

\(f\left(x\right)\ge\dfrac{24\left(x^2+5,76\right)}{16x^2+51,84+9x^2+92,16}=\dfrac{24\left(x^2+5,76\right)}{25\left(x^2+5,76\right)}=\dfrac{24}{25}\)

\(f\left(x\right)_{min}=\dfrac{24}{25}\) khi \(16x^2+51,84=9x^2+92,16\Leftrightarrow x^2=\dfrac{144}{25}\)

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