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\(\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)
ĐKXĐ : x khác 1 , x lớn hơn hoặc bằng 0
\(=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{\sqrt{x}\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\sqrt{x}-1}{1}=\frac{x+2}{\sqrt{x}}\)
b/ \(A=2=\frac{x+2}{\sqrt{x}}\)
\(\Rightarrow2\sqrt{x}=x+2\)
\(\Rightarrow x-2\sqrt{x}+2=0\)
\(\Rightarrow x-2\sqrt{x}+1+1=0\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2+1=0\)
\(\Rightarrow\left(\sqrt{x}-1\right)^2=-1\)
mà\(\left(\sqrt{x}-1\right)^2\ge0\)(ko thỏa mãn)
P/s ko bik phải làm sai ko mà tính ko ra @*@ bạn xem sai chỗ nào để mik sửa ạ
\(A=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\frac{\left(1-x\right)^2}{2}\)
\(A=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)
\(A=\left(\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\frac{\left(1-x\right)^2}{2}\)
\(A=\frac{2}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}.\frac{\left(1+\sqrt{x}\right)^2\left(\sqrt{x}-1\right)^2}{2}\)
\(A=\sqrt{x}-1\)
ý b,c dễ rồi nha
ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\frac{\left(x-1\right)^2}{2}\)
\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}\left(1-\sqrt{x}\right)\)
\(0< x< 1\Rightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\1-\sqrt{x}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x}\left(1-\sqrt{x}\right)>0\Rightarrow A>0\)
\(A< 0\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)< 0\Leftrightarrow1-\sqrt{x}< 0\Rightarrow x>1\)
\(A>-2\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)+2>0\Leftrightarrow-x+\sqrt{x}+2>0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(2-\sqrt{x}\right)>0\Leftrightarrow2-\sqrt{x}>0\Rightarrow x< 4\)
Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0\le x< 4\\x\ne1\end{matrix}\right.\)
\(A< -2x\Leftrightarrow\sqrt{x}-x< -2x\Leftrightarrow x+\sqrt{x}< 0\) (vô nghiệm \(\forall x\ge0\))
\(A>2\sqrt{x}\Leftrightarrow\sqrt{x}-x>2\sqrt{x}\Leftrightarrow x+\sqrt{x}< 0\) giống như trên
\(A=-x+\sqrt{x}=-x+\sqrt{x}-\frac{1}{4}+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)
\(A_{max}=\frac{1}{4}\) khi \(\sqrt{x}=\frac{1}{2}\Leftrightarrow x=\frac{1}{4}\)