Cho biểu thức D=\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
a) tìm điều kiện
b) rút gọn
c) tìm x sao cho D < -1
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13 tháng 8 2021
\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(x\ge0;x\ne3;x\ne-3;x\ne9;x\ne4\right)\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\right):\left(\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ =\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}:\dfrac{9-x+x-9-x+4\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}\\ =\dfrac{3}{\sqrt{x}-2}\)
Tick hộ nha 😘
18 tháng 9 2021
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 \(x\ne9\)
18 tháng 9 2021
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\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
XO
2 tháng 7 2023
a) ĐKXĐ : \(x\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)
b) \(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right).\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\left(x-2\sqrt{x}+1\right)\)
\(=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
c) Có : \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)
Khi đó B = \(\dfrac{\sqrt{3}-1}{2}-1=\dfrac{\sqrt{3}-3}{2}\)
Y
2 tháng 7 2023
\(a,\) B có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(b,B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}-x}{1+\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(x-1\right)-\left(x-1\right)}{1+\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}-1\)
\(c,x=\dfrac{2-\sqrt{3}}{2}\Rightarrow B=\sqrt{\dfrac{2-\sqrt{3}}{2}}-1\)
\(=\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}.\sqrt{2}}-\sqrt{2}\) (Nhân \(\sqrt{2}\) để khử căn dưới mẫu)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-2\sqrt{2}}{2}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-2\sqrt{2}}{2}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-2\sqrt{2}}{2}\)
\(=\dfrac{\sqrt{3}-1-2\sqrt{2}}{2}\)
24 tháng 11 2021
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
23 tháng 7 2021
d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5\sqrt{x}-6-2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{x-3}\)
\(=\dfrac{3\sqrt{x}}{x-3}\)
f) Ta có: \(\left(\dfrac{3}{\sqrt{1+x}}+\sqrt{1-x}\right):\left(\dfrac{3}{\sqrt{1-x^2}}+1\right)\)
\(=\dfrac{3+\sqrt{1-x^2}}{\sqrt{1+x}}:\dfrac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}\)
\(=\dfrac{\sqrt{1-x^2}}{\sqrt{1+x}}=\sqrt{1-x}\)
23 tháng 5 2022
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: 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