Đáp án:
Giải thích các bước giải:
`K=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}`
ĐK: `x \ge 0, x \ne 1`
`K=\frac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{6\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`K=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`K=\frac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`K=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}`
`K=\frac{\sqrt{x}-1}{\sqrt{x}+1}`
`K \le 1/2`
`⇔ \frac{\sqrt{x}-1}{\sqrt{x}+1} \le 1/2`
`⇔ \frac{\sqrt{x}-1}{\sqrt{x}+1} - 1/2 \le 0`
`⇔ \frac{2(\sqrt{x}-1)}{2(\sqrt{x}+1)}-\frac{\sqrt{x}+1}{2(\sqrt{x}+1)} \le 0`
`⇔ \frac{\sqrt{x}-3}{2(\sqrt{x}+1)} \le 0`
Do `x \ge 0⇒\sqrt{x} \ge 0⇒\sqrt{x}+1 \ge 1 ∀x`
`⇔ \sqrt{x}-3 \le 0`
`⇔ \sqrt{x} \le 3`
`⇔ x \le 9` kết hợp ĐKXĐ
Vậy `0 \le x \le 9, x \ne 1` thì `K \le 1/2`
`x \in {0;2;3;4;5;6;7;8;9}`
`A=0+2+3+4+5+6+7+8+9=44`
Chọn B
