`a)`\(\left(x+2\right)\sqrt{x-3}=0\)
\(ĐK:x\ge3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\\sqrt{x-3}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3\right\}\)
`b)`\(x+\sqrt{x-2}=2\sqrt{x-1}\)
\(ĐK:x\ge2\)
\(\Leftrightarrow x^2+x-2+2x\sqrt{x-2}=4x-4\)
\(\Leftrightarrow2x\sqrt{x-2}+x^2-3x+2=0\)
\(\Leftrightarrow2x\sqrt{x-2}+\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\sqrt{x-2}\left[2x+\left(x-1\right)\sqrt{x-2}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\2x+\left(x-1\right)\sqrt{x-2}=0\left(vô.lý\right)\end{matrix}\right.\) ( vì \(x\ge2\) )
Vậy \(S=\left\{2\right\}\)
`c)`\(x^2+x+12\sqrt{x+1}=36\)
\(ĐK:x\ge-1\)
\(\Leftrightarrow x^2+x+12\sqrt{x+1}-36=0\)
\(\Leftrightarrow x^2+2x+1-x-1+12\sqrt{x+1}-36=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x+1-12\sqrt{x+1}+36\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(\sqrt{x+1}-6\right)^2=0\)
`@`TH1:\(x+1=\sqrt{x+1}-6\)
\(\Leftrightarrow x+1-\sqrt{x+1}+6=0\)
Đặt \(\sqrt{x+1}=a;a\ge0\)
\(\Leftrightarrow a^2-a+6=0\)
\(\Delta=\left(-1\right)^2-4.6=-23< 0\)
`->` pt vô nghiệm
`@`TH2:\(x+1=6-\sqrt{x+1}\)
\(\Leftrightarrow x+1+\sqrt{x+1}-6=0\)
Đặt \(\sqrt{x+1}=a;a\ge0\)
\(\Leftrightarrow a^2+a-6=0\)
\(\Delta=1^2-4.\left(-6\right)=25>0\)
\(\rightarrow\left\{{}\begin{matrix}x=\dfrac{-1+\sqrt{25}}{2}=2\left(tm\right)\\x=\dfrac{-1-\sqrt{25}}{2}=-3\left(ktm\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x+1}=2\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
Vậy \(S=\left\{3\right\}\)
`d)`\(\sqrt{5-x}=2x-7\)
\(ĐK:\dfrac{7}{2}\le x\le5\)
\(\Leftrightarrow\left(\sqrt{5-x}\right)^2=\left(2x-7\right)^2\)
\(\Leftrightarrow5-x=4x^2-28x+49\)
\(\Leftrightarrow4x^2-27x+44=0\)
\(\Delta=\left(-27\right)^2-4.4.44=25>0\)
\(\rightarrow\left\{{}\begin{matrix}x=\dfrac{27+5}{8}=4\left(tm\right)\\x=\dfrac{27-5}{8}=\dfrac{11}{4}\left(ktm\right)\end{matrix}\right.\)
Vậy \(S=\left\{4\right\}\)
`e)`\(x^2+4x+5=2\sqrt{2x+3}\)
\(ĐK:x\ge-\dfrac{3}{2}\)
\(\Leftrightarrow x^2+2x+3-2\sqrt{2x+3}+1+2x+1=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1\right\}\)
a: \(\left(x+2\right)\cdot\sqrt{x-3}=0\)
=>x-3=0 hoặc x+2=0
=>x=3
d: \(\sqrt{5-x}=2x-7\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =5\\\left(2x-7-5+x\right)\left(2x-7+5-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =5\\\left(3x-12\right)\left(x-2\right)=0\end{matrix}\right.\Leftrightarrow x=4\)
