căn(2x+8) - căn bậc 3(2x-9)=5

`sqrt{2x + 8} -`$\sqrt[3]{2x - 9}$ `= 5`

ĐKXĐ: `x >= - 4`

Đặt: `sqrt{2x + 8} = a` `(a >= 0)`

$\sqrt[3]{2x - 9}$ `= b`

`=> a^2 - b^3 = 2x + 8 - (2x - 9) = 17`

Ta có hệ pt sau:

$\begin{cases} a - b = 5\\a^2 - b^3 = 17\\ \end{cases}$

`<=>` $\begin{cases} a = 5 + b\\a^2 - b^3 = 17\\ \end{cases}$

`=> (b + 5)^2 - b^3 = 17`

`<=> b^3 - b^2 - 10b - 8 = 0`

`<=> b^3 + 2b^2 - 3b^2 - 6b - 4b - 8 = 0`

`<=> b^2(b + 2) - 3b(b + 2) - 4(b + 2) = 0`

`<=> (b + 2)(b^2 - 3b - 4) = 0`

`<=> (b + 2)(b^2 - 4b + b - 4) = 0`

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

`+)` `TH_1: b = - 2`

`=>` $\sqrt[3]{2x - 9}$ `= - 2`
`<=> x = 1/2` `(TM)`

`+)` `TH_2: b = 4`

`=>` $\sqrt[3]{2x - 9}$ `= 4`

`<=> x = 73/2` `(TM)`

`+)` `TH_3: b = - 1`
`=>` $\sqrt[3]{2x - 9}$ `= - 1`
`<=> x = 4` `(TM)`