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A=(\(3\sqrt{3}-2\sqrt{3}+6\)).\(\sqrt{3}-4\sqrt{3}\)
=\(\sqrt{3}\left(3-2+2\sqrt{3}\right)\).\(\sqrt{3}-4\sqrt{3}\)
=3(\(3-2+2\sqrt{3}\))-4\(\sqrt{3}\)
=3+2\(\sqrt{3}\)
đặt \(\sqrt{x-2}\)=a \(\sqrt{y-3}\)=b
2a+3b=14
\(\left\{{}\begin{matrix}2a+3b=14\\a+b=5\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=5-b\\2.\left(5-b\right)+3b=14\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}a=5-b\\10-2b+3b=14\end{matrix}\right.\)\(\left\{{}\begin{matrix}a=5-b\\b=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=5-4\\b=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}a=1\\b=4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=4\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\sqrt{x-2}=\sqrt{1}\\\sqrt{y-3}=\sqrt{16}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x-2=1\\y-3=16\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=3\\y=19\end{matrix}\right.\)
⇒phương trình có 2 no (x,y)=(3, 19)
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)