Mình k chép lại đề bài nữa nhé
a, Rút gọn:
\(P=\left(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)\(P=\left(\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{x-1}{\sqrt{x}-1}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(P=\left(\dfrac{x-\sqrt{x}+1-x+1}{\sqrt{x}-1}\right):\left(\dfrac{x-\sqrt{x}+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(P=\dfrac{2-\sqrt{x}}{\sqrt{x}-1}:\dfrac{x}{\sqrt{x}-1}\)
\(P=\dfrac{2-\sqrt{x}}{x}\)
b, Để P= 3
\(\Leftrightarrow\dfrac{2-\sqrt{x}}{x}=3\)
\(\Leftrightarrow2-\sqrt{x}=3x\)
\(\Leftrightarrow2-\sqrt{x}-3x=0\)
\(\Leftrightarrow3x+\sqrt{x}-2=0\)
\(\Leftrightarrow3x+3\sqrt{x}-2\sqrt{x}-2=0\)
\(\Leftrightarrow3\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)
\(TH1:3\sqrt{x}-2=0\Leftrightarrow\sqrt{x}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{4}{9}\left(TM\right)\)
\(TH2:\sqrt{x}+1=0\Leftrightarrow\sqrt{x}=-1\Rightarrow v\text{ô}l\text{í}\)
vậy KL
xong nhé