\(a,3\sqrt{x}-7=0\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=7\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{3}\\ \Leftrightarrow x=\dfrac{49}{9}\left(tmdk\right)\)
Vậy \(S=\left\{\dfrac{49}{9}\right\}\)
\(b,\sqrt{x-2}+\sqrt{4x-8}=3\left(dk:x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}=3\\ \Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}=3\\ \Leftrightarrow3\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tmdk\right)\)
Vậy \(S=\left\{3\right\}\)
a: =>3*căn x=7
=>căn x=7/3
=>x=49/9
b: =>3*căn x-2=3
=>căn x-2=1
=>x-2=1
=>x=3
`a, 3 sqrt x - 7 = 0`
`<=> 3 sqrt x = 7`
`<=> sqrt x = 7/3`
`<=> x = 49/9`.
Vậy `x = 49/9`
`b, sqrt(x-2) + sqrt(4x-8) = 3`
`<=> sqrt(x-2) + 2 sqrt(x-2) = 3`
`<=> 3 sqrt(x-2) = 3`
`<=> sqrt(x-2)=1`
`<=> x-2=1`
`<=> x = 3`
Vậy `x = 3`