\(\sqrt{x+4}+\sqrt{x-1}=2\left(x\ge1\right)\)
Đặt: \(\left\{{}\begin{matrix}a=\sqrt{x+4}\\b=\sqrt{x-1}\end{matrix}\right.\left(a,b\ge0\right)\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2=x+4\\b^2=x-1\end{matrix}\right.\)
Ta có: \(\left\{{}\begin{matrix}a+b=2\\a^2-b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\\left(2-b\right)^2-b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\4-4b+b^2-b^2=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\-1-4b=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2-b\\b=-\dfrac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{9}{4}\\b=-\dfrac{1}{4}\left(l\right)\end{matrix}\right.\)
Vậy \(ptvn\)
\(\sqrt{x+4}+\sqrt{x-1}=2\) ( ĐK : \(x\ge1\) )
\(\Leftrightarrow\left(\sqrt{x+4}+\sqrt{x-1}\right)^2=4\)
\(\Leftrightarrow x+4+2\sqrt{\left(x+4\right)\left(x-1\right)}+x-1=4\)
\(\Leftrightarrow2\sqrt{\left(x+4\right)\left(x-1\right)}=1-2x\) ( ĐK : \(x\le\dfrac{1}{2}\) )
\(\Leftrightarrow\sqrt{\left(x+4\right)\left(x-1\right)}=\dfrac{1-2x}{2}\)
\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=\dfrac{4x^2-4x+1}{4}\)
\(\Leftrightarrow x^2+3x-4=\dfrac{4x^2-4x+1}{4}\)
\(\Leftrightarrow4x^2+12x-16=4x^2-4x+1\)
\(\Leftrightarrow16x=17\)
\(\Leftrightarrow x=\dfrac{17}{16}\) ( Loại )
Vậy PTVN
