a)
\(\sqrt{1-x}+\sqrt{4+x}=3\) \(\left(-4\le x\le1\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{4+x}=b\end{matrix}\right.\)\(\left(a,b\ge0\right)\), ta có hpt:
\(\left\{{}\begin{matrix}a+b=3\left(1\right)\\a^2+b^2=5\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow a^2+2ab+b^2=9\)
\(\Leftrightarrow5+2ab=9\)
\(\Leftrightarrow ab=2\). Thay \(b=3-a\) vào, ta có:
\(\Rightarrow a\left(3-a\right)=2\)
\(\Leftrightarrow a^2-3a+2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\) (nhận)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x}=1\\\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) (nhận)
b)
\(x^2+4x+5=2\sqrt{2x+3}\)
\(\Leftrightarrow\left(x+2\right)^2+1=2\sqrt{2\left(x+2\right)-1}\). Đặt \(x+2=t\), ta có:
\(\Rightarrow t^2+1=2\sqrt{2t-1}\)
\(\Leftrightarrow t^4+2t^2+1=8t-4\)
\(\Leftrightarrow\left(t^2+2t+5\right)\left(t-1\right)^2=0\)
\(\Leftrightarrow t=1\) vì \(t^2+2t+5=\left(t+1\right)^2+4\ge4>0\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\)
3/ Ta có:
\(2x\sqrt{y-4}=\sqrt{4x}.\sqrt{xy-4x}\le\dfrac{4x+xy-4x}{2}=\dfrac{xy}{2}\left(1\right)\)
Tương tự ta có:
\(2y\sqrt{x-4}\le\dfrac{xy}{2}\left(2\right)\)
Cộng (1) và (2) vế theo vế ta được:
\(2\left(x\sqrt{y-4}+y\sqrt{x-4}\right)\le xy\)
Dấu = xảy ra khi \(x=y=8\)
