a, ĐKXĐ :\(\left\{{}\begin{matrix}x\ge-3\\x\le6\end{matrix}\right.\)
=> \(-3\le x\le6\)
Ta có : \(\sqrt{x+3}+\sqrt{6-x}-\sqrt{\left(x+3\right)\left(6-x\right)}=3\)
- Đặt \(\sqrt{x+3}=a,\sqrt{6-x}=b\left(a,b\ge0\right)\)
=> \(a^2+b^2=x+3+6-x=9\)
Ta có : \(a+b-ab=3\)
=> \(a+b=ab+3\)
Ta có : \(\left(a+b\right)^2-2ab=9\)
=> \(\left(ab+3\right)^2-2ab=9\)
=> \(a^2b^2+6ab+9-2ab=9\)
=> \(a^2b^2+4ab=0\)
=> \(ab\left(ab+4\right)=0\)
=> \(\left[{}\begin{matrix}ab=0\\ab=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}a+b=3\\a+b=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}a=-b+3\\a=-b-1\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b\left(-b+3\right)=0\\b\left(-b-1\right)=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b\left(-b+3\right)=0\\b^2+b-4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}b=0\left(TM\right)\\b=3\left(TM\right)\\b=\frac{-1-\sqrt{17}}{2}\left(KTM\right)\\b=\frac{-1+\sqrt{17}}{2}\left(TM\right)\end{matrix}\right.\) => \(\left[{}\begin{matrix}6-x=0\\6-x=9\\6-x=\frac{9-\sqrt{17}}{2}\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=6\\x=-3\\x=\frac{3+\sqrt{17}}{2}\end{matrix}\right.\) ( TM )
=> \(\left[{}\begin{matrix}a=3\\a=0\\a=\frac{7-\sqrt{17}}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x+3=0\\x+3=9\\x+3=\frac{33-7\sqrt{17}}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=6\\x=\frac{27-7\sqrt{17}}{2}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\)( TM )
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b, ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ge0\\x+3\ge0\\4-2x\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge1\\x\ge-3\\x\le2\end{matrix}\right.\)
=> \(1\le x\le2\)
Ta có : \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x+3\right)}=4-2x\)
- Đặt \(\sqrt{x-1}=a,\sqrt{x+3}=b\left(a,b\ge0\right)\)
=> \(a^2+b^2=\left(a+b\right)^2-2ab=x-1+x+3=2x+2\)
Ta được: \(a+b+2ab=4-2x=-2x-2+6=-\left(2x+2\right)+6=-a^2-b^2+6\)
=> \(a^2+2ab+b^2=-a-b=\left(a+b\right)^2=-\left(a+b\right)\)
=> \(\left(a+b\right)^2+\left(a+b\right)=0\)
=> \(\left(a+b\right)\left(a+b+1\right)=0\)
=> \(\left[{}\begin{matrix}a+b=0\\a+b+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}\sqrt{x-1}+\sqrt{x+3}=0\\\sqrt{x+1}+\sqrt{x+3}=-1\left(VL\right)\end{matrix}\right.\)
Ta thấy : \(\left\{{}\begin{matrix}\sqrt{x-1}\ge0\\\sqrt{x+3}\ge0\end{matrix}\right.\)
=> \(\sqrt{x-1}+\sqrt{x+3}\ge0\forall x\)
- Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=1\left(TM\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)
=> x = 1 .
Vậy ...
