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1) ĐKXĐ của phân thức là : \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x}-3\ne0\\x-9\ne0\\\sqrt{x}+3\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne3\\\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\ne0\\\sqrt{x}\ne-3\left(LĐ\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có : \(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\left(\dfrac{x+3}{x-9}+\dfrac{1}{\sqrt{x}+3}\right)\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\left(\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{x+\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}.\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\)
\(P=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
2) Với \(x=4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)
Do đó : \(P=\dfrac{\sqrt{3}-1+3}{\sqrt{3}-1+1}\)
\(P=\dfrac{\sqrt{3}+2}{\sqrt{3}}=\dfrac{3+2\sqrt{3}}{3}\)
3) Xét hiệu của : P với 3
\(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-3\)
\(=\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\)
Ta thấy : \(\sqrt{x}+1\ge1;-2\sqrt{x}\le0\)
\(\Rightarrow\dfrac{-2\sqrt{x}}{\sqrt{x}+1}\le0\)
\(\Rightarrow P\le3\)
Dấu bằng xảy ra : \(\Leftrightarrow x=0\). Thế lại ta thấy ktm nên P<3
mk nghỉ ở giữa 2 ngoặc là dấu chia mới đúng chứ :
đk : \(x\ge0;x\ne9\)
\(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\right)\)
\(D=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-\left(3x+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(D=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(D=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
Câu 2:
a, ĐKXĐ: x\(\ge\)0; x\(\ne\)\(\pm\)1
B=
\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-2.2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{-4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ =\dfrac{4\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}}\\ =-\dfrac{4}{\sqrt{x}-1}\)
A=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
ĐKXĐ :x\(\ne\)9,x\(\ge\)0
<=> \(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}\)
=\(\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
=\(\dfrac{3\sqrt{x}-9}{x-9}\)=\(\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3}{\sqrt{x}+3}\)
ta có : A=1/3 => \(\dfrac{3}{\sqrt{x}+3}=\dfrac{1}{3}=>\sqrt{x}+3=9\)
=> x=36
vậy giá trị của x=36 khi A=1/3
B=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{x-1}\right):\dfrac{1}{\sqrt{x}+1}\)
ĐKXĐ: x\(\ne\)1 ,x\(\ge\)0
<=> \(\dfrac{\sqrt{x}+1-\sqrt{x}}{x-1}:\dfrac{1}{\sqrt{x}+1}\)
=\(\dfrac{1}{x-1}:\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x-1}=\dfrac{1}{\sqrt{x}-1}\)
ta có : B<0 =>\(\dfrac{1}{\sqrt{x}-1}< 0\)
=> 1< \(\sqrt{x}-1\)=> \(\sqrt{x}\)>2=>x>4
vậy x>4 khi B<0
\(ĐKXĐ:x\ne9\Rightarrow A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9} =\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{3x+9}{x-9}=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3}{\sqrt{x}+3}\)
\[\begin{array}{l}
Q = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}\left( {\frac{{\sqrt x + 1}}{{\sqrt x - 1}} - \frac{{\sqrt x - 1}}{{\sqrt x + 1}}} \right)\\
Q = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}.\frac{{{{\left( {\sqrt x + 1} \right)}^2} - {{\left( {\sqrt x - 1} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
Q = {\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)^2}.\frac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
Q = \frac{{4\sqrt x {{\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)}^2}}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
Q = \frac{{4\sqrt x {{\left( {\frac{{x - 1}}{{2\sqrt x }}} \right)}^2}}}{{x - 1}}\\
Q = \frac{{\sqrt x .\frac{{{{\left( {x - 1} \right)}^2}}}{x}}}{{x - 1}}\\
Q = \frac{{x\sqrt x - \sqrt x }}{x}
\end{array}\]
a: ĐKXĐ: x>0; x<>4
\(P=\left(2-\sqrt{x}+2\right)\cdot\dfrac{1}{\sqrt{x}-2}=\dfrac{4-\sqrt{x}}{\sqrt{x}-2}\)
b: P=2/3
=>(4-căn x)/(căn x-2)=2/3
=>2căn x-4=12-3căn x
=>5căn x=16
=>x=256/25
c: Khi x=8-2căn 7 thì \(P=\dfrac{4-\sqrt{7}+1}{\sqrt{7}-1-2}=\dfrac{5-\sqrt{7}}{\sqrt{7}-3}=-4-\sqrt{7}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-3x+6\sqrt{x}-11-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
a) DKXD
\(\left\{{}\begin{matrix}x\ge0\\2-\sqrt{x}\ne0\\1-\sqrt{x}\ne0\\x-3\sqrt{x}+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\\\sqrt{x}\ne1\\\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\ne0\end{matrix}\right.\) \(\)
\(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne1\\x\ne1;x\ne4\end{matrix}\right.\)
Vậy điều kiện xác định của biểu thức là x\(\ge\) 0;x\(\ne\) 1;x\(\ne\)4
b) với \(x\ge0;x\ne1;x\ne4\) ta có
A=\(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}-\dfrac{2\sqrt{x}-1}{1-\sqrt{x}}+\dfrac{2-x}{x-3\sqrt{x}+2}\)
= \(\dfrac{(\sqrt{x}-3)\left(1-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(1-\sqrt{x}\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(1-\sqrt{x}\right)}\)
\(+\dfrac{2-x}{\left(2-\sqrt{x}\right)\left(1-\sqrt{x}\right)}\)
=\(\dfrac{-x+4\sqrt{x}-3-4\sqrt{x}+2+2x-\sqrt{x}+2-x}{\left(2-\sqrt{x}\right)\left(1-\sqrt{x}\right)}\)
= \(\dfrac{1-\sqrt{x}}{\left(2-\sqrt{x}\right)\left(1-\sqrt{x}\right)}\)=\(\dfrac{1}{2-\sqrt{x}}\)
Vậy với \(x\ge0;x\ne1;x\ne4\) thì A=\(\dfrac{1}{2-\sqrt{x}}\)
a) Để biểu thức E được xác định thì \(\left\{{}\begin{matrix}\sqrt{x}\ge0\\9x-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x>0\\x\ne\dfrac{1}{9}\end{matrix}\right.\)
b) \(E=\left(1-\dfrac{2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{9x-1}\right):\left(\dfrac{9\sqrt{x}+6}{3\sqrt{x}+1}-3\right)=\left[\dfrac{3\sqrt{x}+1-2\sqrt{x}}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\dfrac{9\sqrt{x}+6-9\sqrt{x}-3}{3\sqrt{x}+1}\right)=\left[\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\dfrac{3}{3\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\left(1+\dfrac{1}{3\sqrt{x}-1}\right).\dfrac{3\sqrt{x}+1}{3}=\dfrac{\sqrt{x}+1}{3\sqrt{x}+1}.\dfrac{3\sqrt{x}+1}{3}.\dfrac{3\sqrt{x}}{3\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{3\sqrt{x}-1}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)