$#nobody$

`sqrtx + sqrt[2 - x] = 1(x \in [0 ; 1])`

`<=> sqrt[2 - x] = 1 - sqrtx`

`<=> 2 - x = 1 - 2sqrtx + x`

`<=> 2x - 2sqrtx - 1 = 0(sqrtx \in [0 ; 1])`

PT trên có:

`\Delta' = (- 1)^2 - 2.(- 1) = 3 > 0`

Do đó,pt trên có `2` nghiệm pb:

`(sqrtx)_1 = (1 + sqrt3)/2 (ktm) ; (sqrtx)_2 = (1 - sqrt3)/2 (ktm)`

Vậy : PT vô n

`sqrt[10 - x] + sqrt[x + 3] = 5(x \in [- 3 ; 10])`

`=> sqrt[x + 3] = 5 - sqrt[10 - x]`

`<=> x + 3 = 25 - 10sqrt[10 - x] + 10 - x`

`=> 2x + 10sqrt[10 - x] - 32 = 0`

`=> x + 5sqrt[10 - x] - 16 = 0`

`=> - x - 5sqrt[10 - x] + 16 = 0`

`=> 10 - x - 5sqrt[10 - x] + 6 = 0`

`=> (sqrt[10 - x] - 3)(sqrt[10 - x] - 2) = 0`

`+) sqrt[10 - x] - 3 = 0 <=> sqrt[10 - x] = 3 <=> 10 - x = 9 <=> x = 1(tm)`

`+) sqrt[10 - x] - 2 = 0 <=> sqrt[10 - x] = 2 <=> 10 - x = 4 <=> x = 6(tm)`

Vậy : ..

`sqrt[x + 1] + sqrt[x + 6] = 5(x \in [- 1 ; 21])`

`<=> sqrt[x + 1] = 5 - sqrt[x + 6]`

`<=> x + 1 = 5 - 10sqrt[x + 6] + x + 6`

`<=> 10 - 10sqrt[x + 6] = 0`

`<=> 10sqrt[x + 6] = 10`

`<=> sqrt[x + 6] = 1`

`<=> x + 6 = 1`

`<=> x = - 5(ktm)`

Vậy : pt vô n

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