\(\left(1-\frac{x}{x-\sqrt{x}+1}\right):\frac{x+2\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\frac{x-\sqrt{x}+1-x}{x-\sqrt{x}+1}.\frac{\left(\sqrt{x}\right)^3+1}{\left(\sqrt{x}+1\right)^2}\)

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

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

\(a,P=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}-2\)

\(P=\frac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1-2x-2\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(b,P=\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)

\(\sqrt{x}-1\ge-1\)

\(\Leftrightarrow\frac{2}{\sqrt{x}-1}\le-2\)

\(\Leftrightarrow P\le1-2=-1\)

\(c,P=4\)\(\Leftrightarrow\sqrt{x}+1=4\sqrt{x}-4\Leftrightarrow3\sqrt{x}=5\Leftrightarrow x=\frac{25}{9}\)

d, Do câu b, \(P\le-1\)\(\Rightarrow P< -3\)

Tối qua bạn có hỏi câu a và mk đã giải đc là:\(P=\frac{3}{\sqrt{x}-1}\)

b)\(P=1\Rightarrow\frac{3}{\sqrt{x}-1}=1\)

\(\Leftrightarrow\sqrt{x}-1=3\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=4^2\)

\(\Leftrightarrow x=16\)(TM)

Vậy khi x=16 thì P=1

\(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)

\(ĐK:\hept{\begin{cases}x^2-4x+1\ge0\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le2-\sqrt{3}\\x\ge2+\sqrt{3}\\x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge2+\sqrt{3}\\0\le x\le2-\sqrt{3}\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-4x+1}=3\sqrt{x}-\left(x+1\right)\)

Bình phương 2 vế ta có :

\(\hept{\begin{cases}3\sqrt{x}-\left(x+1\right)\ge0\\x^2-4x+1=9x-6\sqrt{x}\left(x+1\right)+x^2+2x+1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3\sqrt{x}-\left(x+1\right)\ge0\\6\sqrt{x}\left(x+1\right)=15x\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3\sqrt{x}-\left(x+1\right)\ge0\\3\sqrt{x}\left(2x+2-5\sqrt{x}\right)=0\end{cases}}\)

\(\Leftrightarrow3\sqrt{x}-\left(x+1\right)\ge0\)và \(\orbr{\begin{cases}\sqrt{x}=0\left(lọai\right)\\2x-5\sqrt{x}+2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x-5\sqrt{x}+2=0\\3\sqrt{x}-\left(x+1\right)\ge0\end{cases}}\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{cases}}\)và \(3\sqrt{x}-\left(x+1\right)\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)

\(\Leftrightarrow2x^3=x^3-3x^2+3x-1\)

\(\Leftrightarrow\left(\sqrt[3]{2}x\right)^3=\left(x-1\right)^3\)

\(\Leftrightarrow\sqrt[3]{2}x=x-1\)

\(\Leftrightarrow x\left(1-\sqrt[3]{2}\right)=1\)

\(\Leftrightarrow x=\dfrac{1}{1-\sqrt[3]{2}}\)