1) Ta có: \(P=\dfrac{1}{\sqrt{x}-1}-\dfrac{x\sqrt{x}-\sqrt{x}}{x+1}\left(\dfrac{1}{x-2\sqrt{x}+1}+\dfrac{1}{1-x}\right)\)

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

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

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

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

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

Để \(P=-\dfrac{2}{5}\) thì \(\dfrac{\sqrt{x}-1}{x+1}=\dfrac{-2}{5}\)

\(\Leftrightarrow-2x-2=5\sqrt{x}-5\)

\(\Leftrightarrow-2x-2-5\sqrt{x}+5=0\)

\(\Leftrightarrow-2x-5\sqrt{x}+3=0\)

\(\Leftrightarrow-2x-6\sqrt{x}+\sqrt{x}+3=0\)

\(\Leftrightarrow-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}+3\right)=0\)

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

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

\(\Leftrightarrow-2\sqrt{x}=-1\)

\(\Leftrightarrow x=\dfrac{1}{4}\)(thỏa ĐK)

Biểu thức gì vậy bạn?

tìm các giá trị nguyên của x để biểu thức P=A.B  nhận giá trị nguyên

\(A=\) \(\dfrac{x+2}{x-5}\)

\(=\dfrac{\left(x-5\right)+7}{x-5}\)

\(=1+\dfrac{7}{x-5}\)

để \(\dfrac{7}{x-5}\) ∈Z thì 7⋮x-5

⇒x-5∈\(\left(^+_-1,^+_-7\right)\)

Còn lại thì bạn tự tính nha

a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)

a.

\(x=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\) \(\Rightarrow\sqrt{x}=\sqrt{5}+1\)

 \(\Rightarrow B=\dfrac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\dfrac{\sqrt{5}}{\sqrt{5}+3}=\dfrac{3\sqrt{5}-5}{4}\)

b.

\(B=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

B nguyên \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2=Ư\left(3\right)\)

Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\Rightarrow x=1\)

c??

Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{4\sqrt{x}-1}{x-4}\right):\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{1}\)

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

Để P nguyên thì \(\sqrt{x}-2\in\left\{-1;1\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{1;3\right\}\)

hay \(x\in\left\{1;9\right\}\)

Mik ko hiểu đề