`6)sqrt(2x + 1) = 2/5` (ĐKXĐ: `2x + 1 >= 0 <=> x >= -1/2`)

`2x + 1 = (2/5)^2`

`2x + 1 = 4/25`

`2x = 4/25 - 1`

`2x = -21/25`

`x = -21/50` (tmđk)

Vậy `x = -21/50`

`7)sqrt(3x + 1/2) = 2/7` (ĐKXĐ: `3x + 1/2 >= 0 <=> x >= -1/6`)

`3x + 1/2 = (2/7)^2`

`3x + 1/2 = 4/49`

`3x = 4/49 - 1/2`

`3x = -41/98`

`x = -41/294` (tmđk)

Vậy `x = -41/294`

`8)sqrt(1 - 1/3x) = 3/4` (ĐKXĐ: `1 - 1/3 x >= 0 <=> x <= 3`)

`1 - 1/3 x = (3/4)^2`

`1 - 1/3x = 9/16`

`1/3 x = 1 - 9/16`

`1/3 x = 7/16`

`x = 7/16 : 1/3`

`x = 21/16` (tmđk)

Vậy `x = 21/16`

`9)sqrt(2x + 1) = 1/5` (ĐKXĐ: `2x + 1 >= 0 <=> x >= -1/2`)

`2x + 1 = (1/5)^2`

`2x + 1 = 1/25`

`2x = 1/25 - 1`

`2x = -24/25`

`x = -12/25` (tmđk)

Vậy `x = -12/25`

`10)9/2 - sqrt(x - 1) = 1/3` (ĐKXĐ: `x - 1 >= 0 <=> x >= 1`)

`sqrt(x - 1) = 9/2 - 1/3`

`sqrt(x - 1) = 25/6`

`x - 1 = (25/6)^2`

`x - 1 = 625/36`

`x = 625/36 + 1`

`x = 661/36` (tmđk)

Vậy `x = 661/36`