a) \(\sqrt{x+3}+\sqrt{x-1}=2\) ( ĐK : \(x\ge1\) )
\(\Leftrightarrow x+3+2\sqrt{\left(x+3\right)\left(x-1\right)}+x-1=4\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(x-1\right)}=1-x\) ( ĐK : \(x\le1\) )
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=\left(1-x\right)^2\)
\(\Leftrightarrow4\left(x-1\right)=0\)
\(\Leftrightarrow x=1\left(TM\right)\)
b) \(\sqrt{5x-1}-\sqrt{x-1}=\sqrt{5x-4}\) ( ĐK : \(x\ge1\) )
\(\Leftrightarrow\sqrt{5x-1}=\sqrt{5x-4}+\sqrt{x-1}\)
\(\Leftrightarrow5x-1=5x-4+2\sqrt{\left(5x-4\right)\left(x-1\right)}+x-1\)
\(\Leftrightarrow4-x=2\sqrt{\left(5x-4\right)\left(x-1\right)}\) ( ĐK : \(x\le4\) )
\(\Leftrightarrow\left(x-4\right)^2=4\left(5x-4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2-8x+16=20x^2-36x+16\)
\(\Leftrightarrow19x^2-28x=0\)
\(\Leftrightarrow x\left(19x-28\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTM\right)\\x=\frac{28}{19}\left(TM\right)\end{matrix}\right.\)
a,ĐKXĐ :\(\left\{{}\begin{matrix}x+3\ge0\\x-1\ge0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge-3\\x\ge1\end{matrix}\right.\)
=> \(x\ge1\)
Vậy ĐKXĐ của phương trình trên là \(x\ge1\)
Ta có : \(\sqrt{x+3}+\sqrt{x-1}=2\)
<=> \(\sqrt{x+3}=2-\sqrt{x-1}\)
<=> \(\left(\sqrt{x+3}\right)^2=\left(2-\sqrt{x-1}\right)^2\)
<=> \(x+3=4-4\sqrt{x-1}+x-1\)
<=> \(x+3-4-x+1=-4\sqrt{x-1}\)
<=> \(-4\sqrt{x-1}=0\)
<=> \(\sqrt{x-1}=0\)
<=> \(x-1=0\)
<=> \(x=1\) ( TM )
Vậy phương trình trên có nghiệm là x = 1 .
b, ĐKXĐ : \(\left\{{}\begin{matrix}5x-1\ge0\\x-1\ge0\\5x-4\ge0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}5x\ge1\\x\ge1\\5x\ge4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x\ge\frac{1}{5}\\x\ge1\\x\ge\frac{4}{5}\end{matrix}\right.\)
=> \(x\ge1\)
Vậy ĐKXĐ của phương trình trên là \(x\ge1\)
Ta có :\(\sqrt{5x-1}-\sqrt{x-1}=\sqrt{5x-4}\)
<=> \(\left(\sqrt{5x-1}-\sqrt{x-1}\right)^2=\left(\sqrt{5x-4}\right)^2\)
<=> \(\left(5x-1\right)-2\sqrt{5x-1}\sqrt{x-1}+\left(x-1\right)=5x-4\)
<=> \(5x-1-2\sqrt{5x-1}\sqrt{x-1}+x-1=5x-4\)
<=> \(5x-1+x-1-5x+4=2\sqrt{5x-1}\sqrt{x-1}\)
<=> \(x+2=2\sqrt{5x-1}\sqrt{x-1}\)
<=> \(\left(x+2\right)^2=\left(2\sqrt{5x-1}\sqrt{x-1}\right)^2\)
<=> \(x^2+4x+4=4\left(5x-1\right)\left(x-1\right)\)
<=> \(x^2+4x+4=4\left(5x^2-x-5x+1\right)\)
<=> \(x^2+4x+4=20x^2-4x-20x+4\)
<=> \(x^2+4x+4-20x^2+4x+20x-4=0\)
<=> \(28x-19x^2=0\)
<=> \(x\left(28-19x\right)=0\)
<=>\(\left\{{}\begin{matrix}x=0\\28-19x=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=0\\x=\frac{28}{19}\end{matrix}\right.\)
Mà \(x\ge1\)
=> \(\left\{{}\begin{matrix}x=0\left(L\right)\\x=\frac{28}{19}\left(TM\right)\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là \(x=\frac{28}{19}\)