Câu 58: B

Câu 59: C

Bài 6

Ta có:

sin²x + cos²x = 1

⇒ cos²x = 1 - sin²x

= 1 - 1/9

= 8/9

⇒ cosx = 2√2/3

⇒ tanx = sinx : cosx

= 1/3 : 22/3

= √2/4

⇒ cotx = 1 : tanx

= 1 : √2/4

= 2√2

\(a,P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\left(x\ge0;x\ne1\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x} +1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

\(---\)

\(b,P< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}< \dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{1}{2}< 0\)

\(\Leftrightarrow\dfrac{2\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{2\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}< 0\)

\(\Leftrightarrow\sqrt{x}-3< 0\left(vì.2\left(\sqrt{x}+1\right)>0\forall x\ge0\right)\)

\(\Leftrightarrow\sqrt{x}< 3\)

\(\Leftrightarrow x< 9\)

Kết hợp với điều kiện của \(x\), ta được:

\(0\le x< 9;x\ne1\) thì \(P< \dfrac{1}{2}\)

#\(Toru\)

Bài 2:

Ta có:

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\) 

A nhận giá trị nguyên khi \(\dfrac{4}{\sqrt{x}-3}\) nguyên:

\(\Rightarrow4\) ⋮ \(\sqrt{x}-3\) 

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)

Mà: \(\sqrt{x}-3\ge-3\)

\(\Rightarrow\sqrt{x}-3\in\left\{1;-1;2;-2;4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{4;2;5;1;7\right\}\)

\(\Rightarrow x\in\left\{16;4;25;1;49\right\}\)

\(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-7}{\sqrt{x}+2}=1-\dfrac{7}{\sqrt{x}+2}\) (ĐK: \(x\ge0\))

Để \(A\) nhận giá trị nguyên thì \(1-\dfrac{7}{\sqrt{x}+2}\) nhận giá trị nguyên

\(\Rightarrow\dfrac{7}{\sqrt{x}+2}\) nhận giá trị nguyên

\(\Rightarrow\sqrt{x}+2\inƯ\left(7\right)\)

\(\Rightarrow\sqrt{x}+2\in\left\{1;7;-1;-7\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{-1;5;-3;-9\right\}\) mà \(\sqrt{x}\ge0\forall x\ge0\)

\(\Rightarrow\sqrt{x}=5\)

\(\Rightarrow x=25\left(tmdk\right)\)

#\(Toru\)