a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow3\sqrt{x-3}=\sqrt{4x-12}+4\)
\(\Leftrightarrow9x-27=4x-12+16+8\sqrt{4x-12}\)
\(\Leftrightarrow5x-31=8\sqrt{4x-12}\) \(\left(x\ge\frac{31}{5}\right)\)
\(\Leftrightarrow25x^2-310x+961=256x-768\)
\(\Leftrightarrow25x^2-566x+1729=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=19\\x=\frac{91}{25}\left(loại\right)\end{matrix}\right.\)
Vậy \(x=19\)
b) ĐK: \(x\ge\frac{7}{2}\)
PT \(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2x-7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x-7\\x-5=7-2x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=4\end{matrix}\right.\)
Vây \(x=4\)
b)\(\sqrt{x^2-10\text{x}+25}+7=2\text{x}\) (ĐKXĐ:x\(\in R\))
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}+7=2\text{x}\)
\(\Leftrightarrow\left|x-5\right|+7=2\text{x}\)(1)
TH1:\(x\ge5\)
(1) \(\Leftrightarrow x-5+7=2x\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\)(không thoả mãn)
TH2: \(x< 5\)
(1)\(\Leftrightarrow-3\text{x}=-12\)
\(\Leftrightarrow x=4\)(thoả mãn)
vậy x=4
a) \(3\sqrt{x-3}-\sqrt{4x-12}=4\) (1)
ĐKXĐ: \(x\ge3\)
(1) \(\Leftrightarrow3\sqrt{x-3}-\sqrt{4\left(x-3\right)}=4\)
\(\Leftrightarrow3\sqrt{x-3}-2\sqrt{x-3}=4\)
\(\Leftrightarrow\sqrt{x-3}=4\)
\(\Leftrightarrow x-3=16\)
\(\Leftrightarrow x=19\)
Vậy S = {19}
b) \(\sqrt{x^2-10x+25}+7=2x\) (2)
ĐKXĐ: \(x\ge\frac{7}{2}\)
(2) \(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2x-7\)
\(\Leftrightarrow\left|x-5\right|=2x-7\) (*)
*) Với \(x\ge5\), ta có:
(*) \(\Leftrightarrow x-5=2x-7\)
\(\Leftrightarrow x-2x=-7+5\)
\(\Leftrightarrow-x=-2\)
\(\Leftrightarrow x=2\) (loại)
*) Với x < 5, ta có:
(*) \(\Leftrightarrow5-x=2x-7\)
\(\Leftrightarrow-x-2x=-7-5\)
\(\Leftrightarrow-3x=-12\)
\(\Leftrightarrow x=4\) (nhận)
Vậy S = {4}
