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a) ĐKXĐ: \(x\ge0;x\ne1\)
b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
Sửa đề: \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{x-1}\)
a: ĐKXĐ: x>=0; x<>1
b: \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{x-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2+\left(\sqrt{x}+1\right)^2-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)
\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{\left(\sqrt{x}-1\right)\cdot\left(2\sqrt{x}-1\right)}{x-1}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
a) ĐKXĐ: \(x\ge0,x\ne1\)
b) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{3\sqrt{x}+1}{\sqrt{x}-1}\)
\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1-3\sqrt{x}-1}{\sqrt{x}-1}\)
\(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{-2\sqrt{x}}{\sqrt{x}-1}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{x-2\sqrt{x}+1-2x-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{-x-4\sqrt{x}+1}{x-1}\)
a/ đkxđ: x > 0; x≠1
b/ \(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right):\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\dfrac{x-1}{2\sqrt{x}}\cdot\dfrac{x\sqrt{x}-2x+\sqrt{x}-x\sqrt{x}-2x-\sqrt{x}}{x-1}\)
\(=\dfrac{-4x}{2\sqrt{x}}=-2\sqrt{x}\)
\(a,ĐK:x>0;x\ne1\\ b,A=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ c,x=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow A=\dfrac{2-1}{2}=\dfrac{1}{2}\)
tìm điều kiện xác định có thể rõ ràng chút được không ạ, chỗ này mình không hiểu lắm ý
Câu 1 chưa rõ đề !
Câu 2 :
a ) ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b ) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
c ) \(P=2\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)
2, a,ĐKXĐ:\(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\left\{{}\begin{matrix}\sqrt{x}-2\ne0\\\sqrt{x}+2\ne0\\4-x\ne0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b,\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)\(P=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
c, P=2\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2\)
\(\Leftrightarrow3\sqrt{x}=2\left(\sqrt{x}+2\right)\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\)
Vậy x=16 thì P có giá trị =2