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a: ĐKXĐ: x>0; x<>9
b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{3\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+2}\)
\(=\dfrac{-3\sqrt{x}}{2\sqrt{x}+2}\)
a , thu gọn
\(A=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}-\dfrac{3x+3}{x-9}\right]:\left[\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right]\)
\(A=\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(A=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(A=-\dfrac{3}{\sqrt{x}+3}\)
b , tự làm
\(a\text{) Để biểu thức xác định }\\ \text{thì }\Rightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(\text{b) }A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\\ =\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}-3}\right)\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ =\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\\ =\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ =\dfrac{-3}{\sqrt{x}+3}\)
\(c\text{) Để }A\le-\dfrac{1}{3}\\ \text{thì }\Rightarrow\dfrac{-3}{\sqrt{x}+3}\le-\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}\ge\dfrac{1}{3}\\ \Rightarrow\dfrac{3}{\sqrt{x}+3}-\dfrac{1}{3}\ge0\\ \Rightarrow\dfrac{9}{3\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}\ge0\\ \Rightarrow\dfrac{\sqrt{x}-6}{\sqrt{x}+3}\le0\\ \Leftrightarrow\sqrt{x}-6\ge0\left(\text{Vì }\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}\ge6\\ \Leftrightarrow x\ge36\)
\(d\text{) Do }\sqrt{x}\ge0\\ \Rightarrow\sqrt{x}+3\ge3\\ \Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge-1\\ \text{Dấu }"="\text{ }xảy\text{ }ra\text{ }khi:\text{ }x=0\)
Vậy..............
\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(x\ge0;x\ne3;x\ne-3;x\ne9;x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\\ =\dfrac{3}{\sqrt{x}-2}\)
Tick hộ nha 😘
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
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
a) ĐKXĐ : \(\)\(x\ge0\)
b) C=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
=\(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
=\(\dfrac{3\sqrt{x}-x+x-9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{2\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
=\(\dfrac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right)\left(\sqrt{3}-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
=\(\dfrac{-3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
c)giả sử C<-1
\(\Leftrightarrow\dfrac{-3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+1< 0\)\(\Leftrightarrow\dfrac{-3x+9\sqrt{x}+2\left(x+\sqrt{x}-6\right)}{2x+2\sqrt{x}-12}< 0\)
\(\Leftrightarrow\dfrac{-x+11\sqrt{x}-12}{2x+2\sqrt{x}-12}< 0\)
\(\Leftrightarrow2x+2\sqrt{x}-12< 0\)
\(\Leftrightarrow\)\(2x+6\sqrt{x}-4\sqrt{x}-12< 0\)\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}+3\right)-4\left(\sqrt{x}+3\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-4< 0\\\sqrt{x}+3< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 4\\\sqrt{x}< -3\left(voly\right)\end{matrix}\right.\)
vậy \(0\le x< 4\)khi C<-1
a)
\(B=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{2\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\\ B=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{9-x}:\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{x-3\sqrt{x}}\\ B=\dfrac{\left[\sqrt{x}\left(3-\sqrt{x}\right)\right].\left[\sqrt{x}\left(3-\sqrt{x}+x+9\right)\right]}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\\ B=\dfrac{2.\left(3-\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}=2\dfrac{\sqrt{x}+2}{3+\sqrt{x}}\)
b)
\(B< 1\Leftrightarrow2\dfrac{\sqrt{x}+2}{3+\sqrt{x}}< 1\\ \Leftrightarrow\dfrac{\sqrt{x}+2}{3+\sqrt{x}}< 1\\ \dfrac{\sqrt{x}+2}{3+\sqrt{x}}-1< 0\\ \dfrac{\sqrt{x}+2-3-\sqrt{x}}{3+\sqrt{x}}< 0\\ \dfrac{-1}{3+\sqrt{x}}< 0\\ \Leftrightarrow3+\sqrt{x}>0\Rightarrow x\ge0\left(thõa\:mãn\right)\)
vậy khi \(x\ge0\) thì B<1
a: ĐKXĐ: x>0; x<>9
b: \(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}}{2\sqrt{x}+4}=\dfrac{-3\sqrt{x}}{2\sqrt{x}+4}\)
c: Để D<-1 thì D+1<0
\(\Leftrightarrow-3\sqrt{x}+2\sqrt{x}+4< 0\)
\(\Leftrightarrow4-\sqrt{x}< 0\)
hay x>16