Với x>0, cho hai biểu thức: \(A=\frac{2+\sqrt{x}}{\sqrt{x}}\)và \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\)
a) Rút gọn B.
b) Tìm x để \(\frac{A}{B}>\frac{3}{2}\)
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23 tháng 7 2018
\(B=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right).\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(B=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
\(B=\frac{-\sqrt{x}-1}{\sqrt{x}}\). Vậy ....
30 tháng 7 2018
=\(\left(\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\):\(\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right)\):\(\left(\frac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)=\(\frac{-3\left(\sqrt{x}+1\right)}{x-9}\).\(\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
=\(\frac{-3}{\sqrt{x}+3}\)
QA
27 tháng 2 2022
Trả lời:
b, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x>0\right)\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c, \(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}>\frac{3}{2}\) \(\left(ĐK:x>0\right)\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}>\frac{3}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)
\(\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Rightarrow2\sqrt{x}+1-3\sqrt{x}>0\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-1\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Vậy \(0< x< 1\) là giá trị cần tìm.
2 tháng 2 2022
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4
a, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\) (ĐKXĐ: \(x>0\))
\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
b, \(\frac{A}{B}=\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)
\(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Leftrightarrow2-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)
Kết hợp với điều kiện \(x>0\)ta có: \(0< x< 4\)
Vậy với \(0< x< 4\)thì \(\frac{A}{B}>\frac{3}{2}\)