Bài 1:
1. a) \(\left(\sqrt{27}-2\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-\sqrt{5}\)
\(=\left(3\sqrt{3}-2\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-\sqrt{5}\)
\(=\left(\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-\sqrt{5}\)
\(=3+3\sqrt{5}-\sqrt{5}\)
\(=3+2\sqrt{5}\)
b) \(\sqrt{\left(\sqrt{28}-9\right)^2}+\dfrac{4}{3-\sqrt{7}}\)
\(=\left|\sqrt{28}-9\right|+\dfrac{4\left(3+\sqrt{7}\right)}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}\)
\(=\left(9-\sqrt{28}\right)+\dfrac{4\left(3+\sqrt{7}\right)}{9-7}\left(\sqrt{28}< 9\right)\)
\(=\left(9-2\sqrt{7}\right)+2\left(3+\sqrt{7}\right)\)
\(=9-2\sqrt{7}+6+2\sqrt{7}\)
\(=15\)
2. a) \(\sqrt{2x-1}=\sqrt{2}-1\left(x\ge\dfrac{1}{2}\right)\)
\(\Leftrightarrow2x-1=\left(\sqrt{2}-1\right)^2\)
\(\Leftrightarrow2x-1=3-2\sqrt{2}\)
\(\Leftrightarrow2x=4-2\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{4-2\sqrt{2}}{2}\)
\(\Leftrightarrow x=2-\sqrt{2}\left(tm\right)\)
b) \(\sqrt{25x-25}-21\sqrt{\dfrac{x-1}{49}}=4\cdot\left(3-\sqrt{x-1}\right)\left(x\ge1\right)\)
\(\Leftrightarrow\sqrt{25\left(x-1\right)}-21\cdot\dfrac{\sqrt{x-1}}{\sqrt{49}}=12-4\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}-21\cdot\dfrac{\sqrt{x-1}}{7}=12-4\sqrt{x-1}\)
\(\Leftrightarrow5\sqrt{x-1}-3\sqrt{x-1}=12-4\sqrt{x-1}\)
\(\Leftrightarrow2\sqrt{x-1}+4\sqrt{x-1}=12\)
\(\Leftrightarrow6\sqrt{x-1}=12\)
\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=4+1\)
\(\Leftrightarrow x=5\left(tm\right)\)
Bài II:
1: Khi x=36 thì \(A=\dfrac{6+4}{6+2}=\dfrac{10}{8}=\dfrac{5}{4}\)
2: \(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}+2}{x+16}\)
\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{\sqrt{x}+2}{x-16}\)
3: Đặt P=B(A-1)
\(=\dfrac{\sqrt{x}+2}{x-16}\cdot\left(\dfrac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)\)
\(=\dfrac{\sqrt{x}+2}{x-16}\cdot\dfrac{\sqrt{x}+4-\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{2}{x-16}\)
Để P là số nguyên thì \(2⋮x-16\)
=>\(x-16\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{17;15;14;18\right\}\)
