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a, Với \(x>0;x\ne1\)
\(P=\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}}\right)^2\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\)
\(=\left(\frac{x-1}{2\sqrt{x}}\right)^2\left(\frac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\right)\)
\(=\frac{x^2-2x+1}{4x}.\frac{-4\sqrt{x}}{x-1}=\frac{1-x}{\sqrt{x}}\)
Thay x = 4 => \(\sqrt{x}=2\)vào P ta được :
\(\frac{1-4}{2}=-\frac{3}{2}\)
c, Ta có : \(P< 0\Rightarrow\frac{1-x}{\sqrt{x}}< 0\Rightarrow1-x< 0\)vì \(\sqrt{x}>0\)
\(\Rightarrow-x< -1\Leftrightarrow x>1\)
a, Với x > 0
\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1}{x+\sqrt{x}}=\frac{x-1+1}{x+\sqrt{x}}=\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
b, Ta có : \(A>\frac{2}{3}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{2}{3}>0\Leftrightarrow\frac{3\sqrt{x}-2\sqrt{x}-2}{3\left(\sqrt{x}+1\right)}>0\)
\(\Rightarrow\sqrt{x}-2>0\Leftrightarrow x>4\)
c, \(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+1}.\frac{\sqrt{x}+3}{2\sqrt{x}}=\frac{\sqrt{x}+3}{2\sqrt{x}+2}=\frac{2\sqrt{x}+6}{2\sqrt{x}+2}=1+\frac{4}{2\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+1}\)
\(\Rightarrow\sqrt{x}+1\inƯ\left(2\right)=\left\{1;2\right\}\)
| \(\sqrt{x}+1\) | 1 | 2 |
| \(\sqrt{x}\) | 0 (loại ) | 1 |
| x | loại | 1 |
Trả lời:
a, \(B=\left(\frac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\left(ĐK:x>0;x\ne1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}\right)^3-1}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)
\(=\left(\frac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}-1-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, \(B< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}< \frac{1}{3}\)
\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
Vì \(-\left(\sqrt{x}-1\right)^2< 0\) với mọi \(x>0;x\ne1\)
\(3\left(x+\sqrt{x}+1\right)>0\) với mọi \(x>0;x\ne1\)
\(\Rightarrow\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\) luôn đúng với mọi \(x>0;x\ne1\)
Vậy \(B< \frac{1}{3}\)
c, \(B=\frac{1}{2\sqrt{x}+1}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{1}{2\sqrt{x}+1}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=x+\sqrt{x}+1\)
\(\Leftrightarrow2x+\sqrt{x}=x+\sqrt{x}+1\)
\(\Leftrightarrow x=1\) (tm)
Vậy x = 1 là giá trị cần tìm