Đáp án:
1. Ta có :
`\sqrt{9x^2 - 12x + 4} - 2016 = 1`
`<=> \sqrt{(3x - 2)^2} = 2017`
`<=> |3x - 2| = 2017`
<=> \(\left[ \begin{array}{l}3x - 2 = 2017\\3x - 2 = -2017\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=673\\x=-2015/3\end{array} \right.\)
Vậy `S = {673 ; -2015/3}`
2.
`ĐKXĐ : x ne 9`
`x ≥ 0`
a,Ta có :
`P = (2\sqrt{x})/(\sqrt{x} + 3) + (\sqrt{x})/(\sqrt{x} - 3) + (3x + 9)/(9 - x)`
`= [2\sqrt{x}(\sqrt{x} - 3)]/[(\sqrt{x} + 3)(\sqrt{x} - 3)] + [\sqrt{x}(\sqrt{x} + 3)]/[(\sqrt{x} - 3)(\sqrt{x} + 3)] - (3x + 9)/(x - 9)`
`= (2x - 6\sqrt{x})/(x - 9) + (x + 3\sqrt{x})/(x - 9) - (3x + 9)/(x - 9)`
`= (2x - 6\sqrt{x} + x + 3\sqrt{x} - 3x - 9)/(x - 9)`
`= (-3\sqrt{x} - 9)/(x - 9)`
`= [-3(\sqrt{x} + 3)]/[(\sqrt{x} - 3)(\sqrt{x} + 3)]`
`= (-3)/(\sqrt{x} - 3)`
b, Để `P ≥ 0`
`<=> (-3)/(\sqrt{x} - 3) ≥ 0`
mà `-3 < 0`
`<=> \sqrt{x} - 3 < 0`
`<=> \sqrt{x} < 3`
`<=> 0 ≤ x < 9`
Giải thích các bước giải:
