`a,\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{2x-2}` `(đk: x ≥ 1)`
`⇔(\sqrt{5x-1})^2=(\sqrt{3x-2}+\sqrt{2x-2})^2`
`⇔5x-1=3x-2+2\sqrt{(3x-2)(2x-2)}+2x-2`
`⇔2\sqrt{(3x-2)(2x-2)}=3`
`⇔\sqrt{(3x-2)(2x-2)}=3/2`
`⇔(\sqrt{(3x-2)(2x-2)})^2=(3/2)^2`
`⇔(3x-2)(2x-2)=9/4`
`⇔6x^2-10x+4=9/4`
`⇔6x^2-10x+7/4=0`
`⇔24x^2-40x+7=0`
`⇔\(\left[ \begin{array}{l}x=\dfrac{10+\sqrt{58}}{12}(tm)\\x=\dfrac{10-\sqrt{58}}{12}(ktm)\end{array} \right.\)
`c,\sqrt{2x-1}=\sqrt{3x-2}-\sqrt{5x-1}` `(đk: x ≥ 2/3)`
`⇔\sqrt{2x-1}+\sqrt{5x-1}=\sqrt{3x-2}`
`⇔(\sqrt{2x-1}+\sqrt{5x-1})^2=(\sqrt{3x-2})^2`
`⇔2x-1+2\sqrt{(2x-1)(5x-1)}+5x-1=3x-2`
`⇔4x+2\sqrt{(2x-1)(5x-1)}=0`
`⇔2x+\sqrt{(2x-1)(5x-1)}=0`
`⇔\sqrt{(2x-1)(5x-1)}=-2x`
$\text{Có: x ≥ $\dfrac{2}{3}$}$
`⇒2x > 0 ⇒ -2x < 0`
mà `\sqrt{(2x-1)(5x-1)}>0`
`⇒x∈∅`
