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a, \(x-3\sqrt{x}+2=0\Leftrightarrow x-2\sqrt{x}-\sqrt{x}+2=0\)đk : x >= 0
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\Leftrightarrow x=1;x=4\)
b, \(\sqrt{x^2-1}-\sqrt{x+1}=0\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-\sqrt{x+1}=0\)đk : \(x\ge1\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x-1}-1\right)=0\)
TH1 : \(x=-1\)( loại )
TH2 : \(\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)
c, \(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)đk : x>= -1/2
\(\Leftrightarrow\left(x+2\right)^2-\left(x-1\right)^2-\sqrt{2x+1}=0\)
\(\Leftrightarrow3\left(2x+1\right)-\sqrt{2x+1}=0\Leftrightarrow\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)
TH1 : \(x=-\frac{1}{2}\)
TH2 : \(\sqrt{2x+1}=\frac{1}{3}\Leftrightarrow2x+1=\frac{1}{9}\Leftrightarrow x=\frac{\frac{1}{9}-1}{2}=\frac{-\frac{8}{9}}{2}=-\frac{4}{9}\)
a) ĐK : x \(\ge0\)
\(x-3\sqrt{x}+2=0\)
<=> \(\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)(tm)
b) ĐK \(\hept{\begin{cases}x\ge-1\\x\notin\left\{x\in R|-1< x< 0\right\}\end{cases}}\)
\(\sqrt{x^2-1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x-1}\sqrt{x+1}-\sqrt{x+1}=0\)
<=> \(\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x-1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)(tm)
c) ĐK : \(x\ge-\frac{1}{2}\)
\(x^2+4x+4-\sqrt{2x+1}-\left(x-1\right)^2=0\)
<=> \(6x+3-\sqrt{2x+1}=0\)
<=> \(\sqrt{2x+1}\left(3\sqrt{2x+1}-1\right)=0\)
<=> \(\orbr{\begin{cases}\sqrt{2x+1}=0\\3\sqrt{2x+1}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=-\frac{4}{9}\end{cases}}\)(tm)
a, \(\dfrac{1}{2}\sqrt{x-5}-\sqrt{4x-20+3}=0\left(dkxd:x\ge5\right)\)
\(< =>\dfrac{\sqrt{x-5}}{2}=\sqrt{4x-17}\)
\(< =>\dfrac{x-5}{4}=4x-17\)
\(< =>x-5=16x-68\)
\(< =>15x=68-5=63\)
\(< =>x=\dfrac{63}{15}=\dfrac{21}{5}\)(ktm)
b, \(\sqrt{2x+1}-2\sqrt{x}+1=0\left(dkxd:x\ge0\right)\)
\(< =>\sqrt{2x+1}+1=2\sqrt{x}\)
\(< =>2x+1+1+2\sqrt{2x+1}=4x\)
\(< =>2x-2\sqrt{2x+1}-2=0\)
\(< =>2x+1-2\sqrt{2x+1}+1-4=0\)
\(< =>\left(\sqrt{2x+1}-1\right)^2=4\)
\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}-1=2\\\sqrt{2x+1}-1=-2\end{matrix}\right.\)
\(< =>\left\{{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{2x+1}=-1\left(loai\right)\end{matrix}\right.\)
\(< =>2x+1=9< =>2x=8< =>x=4\)(tmdk)
a) Ta có: \(\dfrac{1}{2}x^2+\dfrac{3}{4}x+1=0\)(1)
\(\Delta=\dfrac{9}{16}-4\cdot\dfrac{1}{2}\cdot1=\dfrac{9}{16}-2=-\dfrac{23}{16}\)
Vì \(\Delta< 0\) nên phương trình (1) vô nghiệm
Vậy: \(S=\varnothing\)
b) Ta có: \(x^2-\left(2+\sqrt{5}\right)x+2\sqrt{5}=0\)(2)
\(\Delta=\left(2+\sqrt{5}\right)^2-4\cdot1\cdot2\sqrt{5}=9+4\sqrt{5}-8\sqrt{5}=9-4\sqrt{5}>0\)
Vì \(\Delta>0\) nên phương trình (2) có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2+\sqrt{5}-\sqrt{9-4\sqrt{5}}}{2\cdot1}=\dfrac{2+\sqrt{5}-\sqrt{5}+2}{2\cdot1}=\dfrac{4}{2}=2\\x_2=\dfrac{2+\sqrt{5}+\sqrt{9-4\sqrt{5}}}{2\cdot1}=\dfrac{2+\sqrt{5}+\sqrt{5}-2}{2\cdot1}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\end{matrix}\right.\)
Vậy: \(S=\left\{2;\sqrt{5}\right\}\)