`a)\sqrt{x}=5(x>=0)`

`<=>x=25(tm)`

Vậy `x=25`

`b)\sqrt{3x}=2(x>=0)`

`<=>3x=4`

`<=>x=4/3(tm)`

Vậy `x=4/3`

`c)3/4\sqrt{2x}=1`

`<=>\sqrt{2x}=4/3(x>=0)`

`<=>2x=16/9`

`<=>x=16/18=8/9(tm)`

Vậy `x=8/9`

`d)\sqrt{x}>2(x>=0)`

`<=>x>4`

Vậy `S={x|x>4}`

`e)2\sqrt{x}<1(x>=0)`

`<=>\sqrt{x}<1/2`

`<=>x<1/4`

`=>S={x|0<=x<1/4}`

`g)\sqrt{x+1}<3(x>=-1)`

`<=>x+1<9`

`<=>x<8`

`=>S={x|-1<=x<8}`

`h)\sqrt{4x-3}=\sqrt{x+2](x>=3/4)`

`<=>4x-3=x+2`

`<=>3x=5`

`<=>x=5/3(tm)`

Vậy `x=5/3`

`i)\sqrt{(3x-1)^2}=\sqrt{(x-2)^2}`

`<=>(3x-1)^2=(x-2)^2`

`<=>(3x-1)^2-(x-2)^2=0`

`<=>(3x-1-x+2)(3x-1+x-2)=0`

`<=>(2x+1)(4x-3)=0`

`<=>`$\left[\begin{matrix} 2x+1=0 \\ 4x-3=0 \end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=\dfrac{-1}{2} \\ x=\dfrac{3}{4} \end{matrix}\right.$

Vậy `x\in{-1/2;3/4}`

a)

$\sqrt{x}$ = 5 ⇔ $(\sqrt{x})^{2}$ = 5 ⇔  x = 5 (ĐKXĐ : x ≥ 0)

b)

$\sqrt{3x}$ = 2 ⇔ $(\sqrt{3x})^{2}$ = 2 ⇔ 3x = 5 ⇔ x = $\frac{5}{3}$ (ĐKXĐ : x ≥ 0)

c)

$\frac{3}{4}$$\sqrt{2x}$ = 1 ⇔ $(\sqrt{2x})$ = $\frac{4}{3}$ ⇔ $(\sqrt{2x})^{2}$ = $\frac{16}{9}$

⇔ 2x = $\frac{16}{9}$ ⇔ x = $\frac{8}{9}$

d)

$\sqrt{x}$ > 2 ⇔ $(\sqrt{x})^{2}$ > 4 ⇔ x > 4 (ĐKXĐ : x ≥ 0)

e)

2$\sqrt{x}$ < 1 ⇔ $\sqrt{x}$ < $\frac{1}{2}$ ⇔ $(\sqrt{x})^{2}$ < $\frac{1}{4}$ ⇔ x < $\frac{1}{4}$

g)

$\sqrt{x+1}$ < 3 ⇔ $(\sqrt{x+1})^{2}$ < 9 ⇔ x + 1 > 9 ⇔ x > 8 (ĐKXĐ : x + 1 ≥ 0)

h)

$\sqrt{4x-3}$ = $\sqrt{x+2}$ ⇔ $(\sqrt{4x-3})^{2}$ = $(\sqrt{x+2})^{2}$ ⇔ 4x-3 = x+2 ⇔ x = $\frac{5}{3}$

i)

$\sqrt{({3x-1})^{2}}$ = $\sqrt{({x-2})^{2}}$ ⇔ $|3x-1|$ = $|x-2|$ 

⇔ 3x - 1 = -(x - 2)    hay     3x - 1 = x - 2

⇔ x = $\frac{5}{4}$  hay  x = $\frac{-1}{2}$