bài 1) ta có : \(\dfrac{\left(5-x\right)\sqrt{5-x}+\left(3-x\right)\sqrt{3-x}}{\sqrt{5-x}+\sqrt{3-x}}=2\) đk : \(x\le3\)
\(\Leftrightarrow\left(5-x\right)\sqrt{5-x}+\left(3-x\right)\sqrt{3-x}=2\left(\sqrt{5-x}+\sqrt{3-x}\right)\)
\(\Leftrightarrow\left(\sqrt{5-x}+\sqrt{3-x}\right)\left(5-x+\sqrt{\left(5-x\right)\left(3-x\right)}+3-x\right)=2\left(\sqrt{5-x}+\sqrt{3-x}\right)\)
\(\Leftrightarrow5-x+\sqrt{\left(5-x\right)\left(3-x\right)}+3-x=2\)
\(\Leftrightarrow\sqrt{\left(5-x\right)\left(3-x\right)}=2x-6\) đk : \(x\ge3\) \(\Rightarrow x=3\)
thử lại ta thấy \(x=3\) là nghiệm
vậy \(x=3\)
bài 2) ta có : \(\sqrt{x-5}-\dfrac{x-14}{3+\sqrt{x-5}}=3\) đk : \(x\ge5\)
\(\Leftrightarrow\dfrac{3\sqrt{x-5}+x-5-x+14}{3+\sqrt{x-5}}=3\)
\(\Leftrightarrow3\sqrt{x-5}+9=9+3\sqrt{x-5}\) (luôn đúng)
vậy \(x\ge5\)