cuuuuuuuuuuuuuuuuuuuuuuu

`color{green}{#adonis}`

Đáp án:

 `S={3}`

Giải thích các bước giải:

Ta có: `sqrt{x-2} - sqrt{4-x} = 2x^2 -5x - 3` `ĐKXĐ: 2<=x<=4`

`<=> (x-2-4+x)/(sqrt{x-2} + sqrt{4-x}) = 2x^2 -6x+x-3`

`<=> (2x-6)/(sqrt{x-2} + sqrt{4-x}) = 2x(x-3)+(x-3)`

`<=> (2(x-3))/(sqrt{x-2} + sqrt{4-x}) - (x-3)(2x+1)=0`

`<=> (x-3)[2/(sqrt{x-2} + sqrt{4-x}) - (2x+1)]=0`

`<=> x-3=0` hoặc `2/(sqrt{x-2} + sqrt{4-x}) - (2x+1)=0`

`<=> x=3` hoặc `(2-(2x+1)(sqrt{x-2} + sqrt{4-x}))/(sqrt{x-2} + sqrt{4-x})=0` `(1)`

Xét `(1)` ta có:

`2-(2x+1)(sqrt{x-2} + sqrt{4-x}) = 0`

`(sqrt{x-2} + sqrt{4-x}) > 1 AA x>=2`

`2x+1 >=5 AA x>=2`

`=> (2x+1)(sqrt{x-2} + sqrt{4-x}) >5`

`-> 2-(2x+1)(sqrt{x-2} + sqrt{4-x}) <0 AA x>=2`

Vậy `(1)` vô nghiệm

Vậy `S={3}`