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Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
a)\(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}.\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\right)\)
=\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b)P=3/2 <=>\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{3}{2}\Leftrightarrow2\sqrt{x}+1=\frac{3}{2}\sqrt{x}+\frac{3}{2}.\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với x=1 thoả nãm yêu cầu
\(a,Đkxđ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x^3}-1\right)}{x+\sqrt{x}+1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=x-\sqrt{x}\)
\(b,P=x-\sqrt{x}=x-\sqrt{x}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Ta có: \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\forall x\ge0\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow x=\frac{1}{4}\)
\(Min_P=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)
c, Đề thiếu không bạn?
Không bn nha