Question

Bài 1: A\(=\frac{3\sqrt{x}-6}{x-2\sqrt{x}}+\frac{\sqrt{x}-3}{\sqrt{x}}-\frac{1}{2-\sqrt{x}}\) và B \(=\frac{\sqrt{x}-2}{\sqrt{x}+9}\)

a, Rút gọn A

b, P= A.B. Tìm các số nguyên x để \(\sqrt{P}< \frac{1}{3}\)

Answer 1

https://i.imgur.com/Kyn6M5n.jpg

Answer 2

https://i.imgur.com/HuIhMUh.jpg

Answer 3

a) Từ đề bài suy ra A = \(\frac{3\sqrt{x}-6}{x-2\sqrt{x}}+\frac{\sqrt{x}-3}{\sqrt{x}}+\frac{1}{\sqrt{x}-2}\) ĐKXĐ của A, B là: x > 0 ; x ≠ 4

A = \(\frac{3\sqrt{x}-6+\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

A = \(\frac{3\sqrt{x}-6+x-5\sqrt{x}+6+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

A = \(\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

A = \(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\) = \(\frac{\sqrt{x}-1}{\sqrt{x}-2}\)

b) Ta có P = A.B = \(\frac{\sqrt{x}-1}{\sqrt{x}-2}.\frac{\sqrt{x}-2}{\sqrt{x}+9}\) = \(\frac{\sqrt{x}-1}{\sqrt{x}+9}\) Suy ra ĐK: x ≥ 1 (x ∈ Z)

\(\sqrt{\frac{\sqrt{x}-1}{\sqrt{x}+9}}< \frac{1}{3}\) ⇔ \(\frac{\sqrt{x}-1}{\sqrt{x}+9}< \frac{1}{9}\) ⇔ \(9\sqrt[]{x}-9< \sqrt{x}+9\) ⇔ \(8\sqrt{x}< 18\) ⇔1 ≤ \(x< \frac{81}{16}\)

Answer 4

https://i.imgur.com/0vsYb6S.jpg