Đáp án:
`ĐKXĐ : x ≥ 3`
Ta có
`2\sqrt{x + 4} - 4\sqrt{2x - 6} = x - 7`
`<=> 2\sqrt{x + 4} - 4\sqrt{2x - 6} - (x - 7) = 0`
`<=> (2\sqrt{x + 4} - 6) - (4\sqrt{2x - 6} - 8) - (x - 7 + 2) = 0`
`<=> 2(\sqrt{x + 4} - 3) - 4(\sqrt{2x - 6} - 2) - (x - 5) = 0`
`<=> 2 . (x + 4 - 9)/(\sqrt{x + 4} + 3) - 4 . (2x - 6 - 4)/(\sqrt{2x - 6} + 2) - (x - 5) = 0`
`<=> [2(x - 5)]/(\sqrt{x + 4} + 3) - [8(x - 5)]/(\sqrt{2x - 6} + 2) - (x - 5) = 0`
`<=> (x - 5)(2/(\sqrt{x + 4} + 3) - 8/(\sqrt{2x - 6} + 2) - 1) = 0`
`+) x - 5 = 0 <=> x = 5`
`+) 2/(\sqrt{x + 4} + 3) - 8/(\sqrt{2x - 6} + 2) - 1 = 0`
`<=> 2(\sqrt{2x - 6} + 2) - 8(\sqrt{x + 4} + 3) - (\sqrt{x + 4} + 3)(\sqrt{2x - 6} + 2) = 0`
`-> Vn_{o}`
Vậy `S = {5}`
Giải thích các bước giải:
