\(\sqrt{30-x}-\sqrt{x-5}=\sqrt{x-13}\left(1\right)\)
ĐKXĐ: \(13\le x\le30\)
\(\left(1\right)\Leftrightarrow\sqrt{30-x}=\sqrt{x-13}+\sqrt{x-5}\)
\(\Leftrightarrow30-x=x-13+x-5+2\sqrt{\left(x-13\right)\left(x-5\right)}\)
\(\Leftrightarrow2\sqrt{\left(x-13\right)\left(x-5\right)}=48-3x\)
\(\Leftrightarrow\left\{{}\begin{matrix}48-3x\ge0\\4\left(x-13\right)\left(x-5\right)=\left(48-3x\right)^2\end{matrix}\right.\)
+) \(48-3x\ge0\Leftrightarrow3x\le48\Leftrightarrow x\le16\)
+) \(4\left(x-13\right)\left(x-5\right)=\left(48-3x\right)^2\)
\(\Leftrightarrow4x^2-72x+260=2304-288x+9x^2\)
\(\Leftrightarrow5x^2-216x+2044=0\)
△' \(=108^2-2044.5=1444>0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\frac{108-\sqrt{1444}}{5}\\x_2=\frac{-108-\sqrt{1444}}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=14\\x_2=\frac{-146}{5}\end{matrix}\right.\)
Đối chiếu đk thì chỉ có \(x=14\)thỏa mãn
Vậy pt có nghiệm là \(x=14\)
Gọn nhẹ hơn 1 chút:
ĐKXĐ:...
\(\Leftrightarrow\sqrt{x-13}-1+\sqrt{x-5}-3+4-\sqrt{30-x}=0\)
\(\Leftrightarrow\frac{x-14}{\sqrt{x-13}+1}+\frac{x-14}{\sqrt{x-5}+3}+\frac{x-14}{4+\sqrt{30-x}}=0\)
\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{\sqrt{x-13}+1}+\frac{1}{\sqrt{x-5}+3}+\frac{1}{4+\sqrt{30-x}}\right)=0\)
\(\Leftrightarrow x=14\)
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