`(\sqrt{x}+2)(3-2\sqrt{x})=5-2x` `ĐK: x >= 0`
`<=>3\sqrt{x}-2x+6-4\sqrt{x}=5-2x`
`<=>\sqrt{x}=1`
`<=>x=1` (t/m)
Vậy `S={1}`
\(đKx\ge0\\ 3\sqrt{x}+6-2x-4\sqrt{x}-5+2x=0\\ -\sqrt{x}+1=0\\ -\sqrt{x}=-1\\ \sqrt{x}=1\\ x=1\left(thoaman\right)\)
\(\Leftrightarrow3\sqrt{x}-2x+6-4\sqrt{x}=5-2x\)
=>-căn x=-1
=>x=1
(√x+2)(3−2√x)=5−2x(x+2)(3-2x)=5-2x ĐK:x≥0ĐK:x≥0
⇔3√x−2x+6−4√x=5−2x⇔3x-2x+6-4x=5-2x
⇔√x=1⇔x=1
⇔x=1⇔x=1 (t/m)
Vậy S={1}
