đk : \(x\ge3\)
\(\left(1\right)\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(\sqrt{x-3}-1\right)=-x^2+5x-4\)
\(\Leftrightarrow\frac{\left(2x+1\right)-3}{\sqrt{2x+1}+3}+\frac{\left(x-3\right)-1}{\sqrt{x-3}+1}=-\left(x-1\right)\left(x-4\right)\)
\(\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{x-3}+1}=\left(-x+1\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}+x-1\right)=0\)
\(\Leftrightarrow\left(x-4\right).f\left(x\right)=0\)
<=> x - 4 = 0 ( vì khji \(x\ge3\)thì \(f\left(x\right)>0\))
<=> x = 4 ( tmđk )
Vậy x = 4 là nghiệm của pt đã cho
đk: \(x\ge3\)
\(PT\Leftrightarrow\left(x^2-5x+4\right)+\left(\sqrt{2x+1}-3\right)+\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-4\right)+\frac{2x-8}{\sqrt{2x+1}+3}+\frac{x-4}{\sqrt{x-3}+1}=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}\right)=0\)
Vì \(x\ge3\) theo đk nên: \(x-1+\frac{2}{\sqrt{2x+1}+3}+\frac{1}{\sqrt{x-3}+1}>0\)
\(\Rightarrow x-4=0\Rightarrow x=4\)(tm)
Vậy x = 4
