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a) điều kiện xác định : \(a\ge0;a\ne1\)
ta có : \(P=\dfrac{3a+\sqrt{9a}-3}{a+\sqrt{a}-2}-\dfrac{\sqrt{a}-2}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+2}-1\)
\(\Leftrightarrow P=\dfrac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\dfrac{\sqrt{a}-2}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}+2}\) \(\Leftrightarrow P=\dfrac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\) \(\Leftrightarrow P=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)để \(\left|P\right|=1\Leftrightarrow\left|\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right|=1\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=1\\\dfrac{\sqrt{a}+1}{\sqrt{a}-1}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-1=0\\\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{\sqrt{a}-1}=0\\\dfrac{2\sqrt{a}}{\sqrt{a}-1}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2=0\left(vôlí\right)\\2\sqrt{a}=0\end{matrix}\right.\Rightarrow a=0\)
vậy \(a=0\)
a: \(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}-3}\)
b: A=1/3
=>\(\dfrac{-3}{\sqrt{x}-3}=\dfrac{1}{3}\)
=>căn x-3=-9
=>căn x=-6(loại)
c: căn x-3>=-3
=>3/căn x-3<=-1
=>-3/căn x-3>=1
Dấu = xảy ra khi x=0
1: \(P=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)
2: Để P nguyên thì \(\sqrt{a}-1+2⋮\sqrt{a}-1\)
\(\Leftrightarrow\sqrt{a}-1\in\left\{1;-1;2\right\}\)
hay \(a\in\left\{4;0;9\right\}\)
\(a,x=16\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)
\(b,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{\sqrt{x}-7}{1-x}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-\dfrac{\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-7\right)}{x-1}\\ =\dfrac{x+4\sqrt{x}-5-\sqrt{x}+7}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(dpcm\right)\)
\(c,\dfrac{4A}{A}\le\dfrac{x}{\sqrt{x}-3}\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}:\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow\dfrac{4\left(\sqrt{x}+2\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\le\dfrac{x}{\sqrt{x}-3}\)
\(\Leftrightarrow4-\dfrac{x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-12-x}{\sqrt{x}-3}\le0\)
\(\Leftrightarrow\) Pt vô nghiệm
Vậy không có giá trị x thỏa yêu cầu đề bài.
a.(ĐKXĐ: \(a\ge0,a\ne\dfrac{1}{9}\))
=> \(A=\left(\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{1+3\sqrt{a}}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\right):\dfrac{3\sqrt{a}+1-3\sqrt{a}+2}{3\sqrt{a}+1}=\dfrac{\left(\sqrt{a}-1\right)\left(1+3\sqrt{a}\right)-3\sqrt{a}+1+8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}.\dfrac{3\sqrt{a}+1}{3}\)
\(=\dfrac{\sqrt{a}+3a-1-3\sqrt{a}-3\sqrt{a}+1+8\sqrt{a}}{3\left(3\sqrt{a}-1\right)}=\dfrac{3\sqrt{a}+3a}{3\left(3\sqrt{a}-1\right)}=\dfrac{3\left(\sqrt{a}+a\right)}{3\left(3\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+a}{3\sqrt{a}-1}\)
b. Để A \(=\dfrac{6}{5}\Leftrightarrow\dfrac{\sqrt{a}+a}{3\sqrt{a}-1}=\dfrac{6}{5}\)
\(\Leftrightarrow5\left(\sqrt{a}+a\right)=6\left(3\sqrt{a}-1\right)\)
\(\Leftrightarrow5\sqrt{a}+5a-18\sqrt{a}+6=0\)
\(\Leftrightarrow5a-13\sqrt{a}+6=0\)
\(\Leftrightarrow5a-10\sqrt{a}-3\sqrt{a}+6=0\)
\(\Leftrightarrow5\sqrt{a}\left(\sqrt{a}-2\right)-3\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(5\sqrt{a}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-2=0\\5\sqrt{a}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=2\\\sqrt{a}=\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\\a=\dfrac{9}{25}\end{matrix}\right.\)(nhận)
Vậy ...
a: \(A=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)
b: \(\sqrt{x}+3>=3\)
=>A<=1
Dấu = xảy ra khi x=0
c: \(P=A:\left(B-1\right)=\dfrac{3}{\sqrt{x}+3}:\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}-2}\)
Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{1;25\right\}\)
DAT P = Q:R \(Q=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(3\sqrt{a}-1\right)}-\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-3\right)\left(3\sqrt{a}-1\right)}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(=\dfrac{\sqrt{a}-1}{3\sqrt{a}-1}-\dfrac{1}{3\sqrt{a}+1}+\dfrac{8\sqrt{a}}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(=\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\)
\(R=1-\dfrac{2\sqrt{a}-a+1}{3\sqrt{a}+1}=\dfrac{a+\sqrt{a}}{3\sqrt{a}+1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{3\sqrt{a}+1}\)
\(\Rightarrow P=Q:R=\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)}{\left(3\sqrt{a}-1\right)\left(3\sqrt{a}+1\right)}\times\dfrac{3\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(P=\dfrac{3}{3\sqrt{a}-1}\)
\(P>\dfrac{3}{\left|1-3\sqrt{5}\right|}\Leftrightarrow\dfrac{3}{3\sqrt{a}-1}>\dfrac{3}{3\sqrt{5-1}}\)
\(3\sqrt{a}-1< 3\sqrt{5}-1\)
\(\Rightarrow0\le\sqrt{a}\le\sqrt{5}\)
\(a=\) 0 ;1 ;2 ;3 ;4
a lớn nhất \(\Rightarrow a\) = 4
Bạn rút gọn được P chưa ?~!