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a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(=\dfrac{-6}{\sqrt{x}+3}\)
b: Để A<-1/2 thì A+1/2<0
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow-12+\sqrt{x}+3< 0\)
=>0<x<81 và x<>9
a: \(A=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}+\sqrt{5}+3-3-\sqrt{5}=1\)
b: \(B=\dfrac{-\sqrt{x}-3+x-3\sqrt{x}-x-9}{x-9}=\dfrac{-4\sqrt{x}-12}{x-9}=\dfrac{-4}{\sqrt{x}-3}\)
Để B>1 thì \(\dfrac{-4-\sqrt{x}+3}{\sqrt{x}-3}>0\)
\(\Leftrightarrow\sqrt{x}-3< 0\)
hay 0<x<9
a: \(P=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b: ĐểP<15/4 thì P-15/4<0
\(\Leftrightarrow4\left(3\sqrt{x}+8\right)-15\left(\sqrt{x}+2\right)< 0\)
=>12 căn +32-15 căn x+30<0
=>-3 căn x<-62
=>căn x>62/3
=>x>3844/9
a:
Sửa đề: \(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(C=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=-\dfrac{3\sqrt{x}}{2\sqrt{x}+4}\)
b: Để C<-1 thì C+1<0
=>-3 căn x+2 căn x+4<0
=>-căn x<-4
=>x>16
1) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\x-9\ne0\\\sqrt{x}-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}=\dfrac{3\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}\)2) Để A=\(\dfrac{5}{6}\) thì \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+3\right)}=\dfrac{5}{6}\Leftrightarrow\left(\sqrt{x}+1\right)6=\left(\sqrt{x}+3\right)5\Leftrightarrow6\sqrt{x}+6=5\sqrt{x}+15\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)
1. Ta có:
\(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)}{3\left(x-9\right)}+\dfrac{1}{3}\)
\(=\dfrac{2x-6\sqrt{x}}{3\left(x-9\right)}+\dfrac{x-9}{3\left(x-9\right)}\)
\(=\dfrac{3x-6\sqrt{x}-9}{3x-27}\)
\(=\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(A=\sqrt{\dfrac{18-3\sqrt{3}}{11}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{\sqrt{11\left(18-3\sqrt{3}\right)}}{11}-\dfrac{\sqrt{6}+\sqrt{2}}{2}\)
\(=\dfrac{2\sqrt{11\left(18-3\sqrt{3}\right)}-11\sqrt{6}-11\sqrt{2}}{22}\)
b: \(=\dfrac{x\sqrt{x}-2x+28-x+16-x-9\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x-9\sqrt{x}+36}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{x-9}{\sqrt{x}+1}\)
Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)
\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3x+9\sqrt{x}}{x-9}=\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-1}\)
Vậy \(P=\dfrac{3\sqrt{x}}{\sqrt{x}-1}\)
Nguyễn Phương Hiền : hì, nhầm, sorry nha ^^!