Đáp án:
a, `ĐKXĐ A : x ≥ 0`
`B : x ≥ 0 ; x ne 9`
b, Với `x = 36`
`=> A = 10/(\sqrt{36} + 8) = 10/(6 + 8) = 10/14 = 5/7`
c, Ta có :
`B = (\sqrt{x})/(\sqrt{x} - 3) + (2\sqrt{x} - 24)/(x - 9)`
`= [\sqrt{x}(\sqrt{x} + 3)]/[(\sqrt{x} - 3)(\sqrt{x} + 3)] + (2\sqrt{x} - 24)/[(\sqrt{x} - 3)(\sqrt{x} + 3)] `
`= (x + 3\sqrt{x})/[(\sqrt{x} - 3)(\sqrt{x} + 3)] + (2\sqrt{x} - 24)/[(\sqrt{x} - 3)(\sqrt{x} + 3)] `
`= (x + 3\sqrt{x} + 2\sqrt{x} - 24)/[(\sqrt{x} - 3)(\sqrt{x} + 3)] `
`= (x + 5\sqrt{x} - 24)/[(\sqrt{x} - 3)(\sqrt{x} + 3)] `
`= [(x - 3\sqrt{x}) + (8\sqrt{x} - 24)]/[(\sqrt{x} - 3)(\sqrt{x} + 3)] `
`= [\sqrt{x}(\sqrt{x} - 3) + 8(\sqrt{x} - 3)]/[(\sqrt{x} - 3)(\sqrt{x} + 3)] `
`= [(\sqrt{x} - 3)(\sqrt{x} + 8)]/[(\sqrt{x} - 3)(\sqrt{x} + 3)]`
`= (\sqrt{x} + 8)/(\sqrt{x} + 3)`
`=> P = A . B = 10/(\sqrt{x} + 8) . (\sqrt{x} + 8)/(\sqrt{x} + 3) = 10/(\sqrt{x} + 3)`
d, Để `P ∈ Z <=> 10/(\sqrt{x} + 3) ∈ Z`
`<=> \sqrt{x} + 3 ∈ Ư(10)`
Do `\sqrt{x} + 3 ≥ 3`
`<=> \sqrt{x} + 3 ∈ {5 ; 10}`
`<=> \sqrt{x} ∈ {2 ; 7}`
`<=> x ∈ {4 ; 49}`
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