Với \(x\ge1\)
\(\sqrt{x+4}-\sqrt{x-1}=1\)
<=>\(\sqrt{x+4}=\sqrt{x-1}+1\)
<=>\(x+4=x-1+1+2\sqrt{x-1}\)
<=>\(2\sqrt{x-1}=4\)
<=>\(\sqrt{x-1}=2\)
<=>\(x-1=4\)
<=>x=5(TM)
bước tương đương thứ 2 bình phương cả 2 vế
đk : x>= 1
\(\sqrt{x+4}-3-\left(\sqrt{x-1}-2\right)=0\)
\(\Leftrightarrow\dfrac{x+4-9}{\sqrt{x+4}+3}-\dfrac{x-1-4}{\sqrt{x-1}+2}=0\Rightarrow x-5=0\Leftrightarrow x=5\left(tm\right)\)
\(dkd:x\ge1\)
\(pt\Leftrightarrow\sqrt{x+4}-3-\sqrt{x-1}+2=0\)
\(\Leftrightarrow\dfrac{x+4-9}{\sqrt{x+4}+3}-\dfrac{x-1-4}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\dfrac{x-5}{\sqrt{x+4}+3}-\dfrac{x-5}{\sqrt{x-1}+2}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\dfrac{1}{\sqrt{x+4}+3}-\dfrac{1}{\sqrt{x-1}+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\\dfrac{1}{\sqrt{x+4}+3}-\dfrac{1}{\sqrt{x-1}+2}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x-1}+2-\sqrt{x+4}-3}{\left(\sqrt{x+4}+3\right)\left(\sqrt{x-1}+2\right)}=0\)
\(\Rightarrow\sqrt{x-1}+2-\sqrt{x+4}-3=0\)
\(\Leftrightarrow\sqrt{x-1}-\sqrt{x+4}-1=0\)
\(\Leftrightarrow\sqrt{x+4}-\sqrt{x-1}+1=0\)
\(do:x\ge1\Rightarrow\sqrt{x+4}-\sqrt{x-1}+1>0\Rightarrow vô-nghiệm\)
\(\Rightarrow S=\left\{5\right\}\)
