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\(E=\left(\frac{a-1}{2\sqrt{a}}\right)\left[\frac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(E=\frac{a-1}{2\sqrt{a}}.\frac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{a-1}\)
\(E=\frac{\left(\sqrt{a}-1+\sqrt{a}+1\right)\left(\sqrt{a}-1-\sqrt{a}-1\right)}{2}=\frac{2\sqrt{a}.-2}{2}=-2\sqrt{a}\)
\(E=-4\Leftrightarrow-2\sqrt{a}=-4\Leftrightarrow\sqrt{a}=2\Leftrightarrow a=4\)(nhận)
a) ĐKXĐ : \(a>0;a\ne1\)
\(Q=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}+2}-\frac{\sqrt{a}-2}{\sqrt{a}-1}\right)\)
\(Q=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\right)\)
\(Q=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(a-1\right)-\left(a-4\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{3}\)
\(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}\)
b) \(Q=\frac{\sqrt{a}+2}{3\sqrt{a}}>2\Rightarrow\sqrt{a}-6\sqrt{a}+2>0\Rightarrow-5\sqrt{a}>-2\Rightarrow0< \sqrt{a}< \frac{2}{5}\)
\(\Rightarrow0< a< \frac{4}{25}\)
1/
a/ ĐKXĐ: \(x\ge0\) và \(x\ne\frac{1}{9}\)
b/ \(P=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}\right]:\left(\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\right)\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}+1\right)\left(3\sqrt{x}-1\right)}.\frac{3\sqrt{x}+1}{3}\)
\(=\frac{3x+3\sqrt{x}}{3\sqrt{x}-1}.\frac{1}{3}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c/ \(P=\frac{6}{5}\Rightarrow\frac{x+\sqrt{x}}{3\sqrt{x}-1}=\frac{6}{5}\Rightarrow6\left(3\sqrt{x}-1\right)=5\left(x+\sqrt{x}\right)\)
\(\Rightarrow5x-13\sqrt{x}+6=0\Rightarrow\left(5\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=\frac{3}{5}\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}}\)
Vậy x = 9/25 , x = 4
1) a) ĐKXĐ : \(0\le x\ne\frac{1}{9}\)
b) \(P=\left(\frac{\sqrt{x}-1}{3\sqrt{x}-1}-\frac{1}{3\sqrt{x}+1}+\frac{8\sqrt{x}}{9x-1}\right):\left(1-\frac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)
\(=\left[\frac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}-\frac{3\sqrt{x}-1}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}+\frac{8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}\right]:\frac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)
\(=\frac{3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}.\frac{3\sqrt{x}+1}{3}=\frac{3x+3\sqrt{x}}{3\left(3\sqrt{x}-1\right)}=\frac{x+\sqrt{x}}{3\sqrt{x}-1}\)
c) \(P=\frac{6}{5}\Leftrightarrow18\sqrt{x}-6=5x+5\sqrt{x}\Leftrightarrow5x-13\sqrt{x}+6=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{25}\\x=4\end{cases}}\)
1.
ĐK \(a\ge0;a\ne1\)
Ta có \(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right).\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\)
\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{a-1}{\sqrt{a}}\)
\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
\(=\frac{4a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}=4a\)
2. Với \(a=\frac{\sqrt{6}}{2+\sqrt{6}}\Rightarrow A=\frac{4\sqrt{6}}{2+\sqrt{6}}\)
Để \(\sqrt{A}>A\Rightarrow\sqrt{4a}>4a\Rightarrow2\sqrt{a}-4a>0\Rightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)>0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{a}>0\\1-2\sqrt{a}>0\end{cases}\Rightarrow\hept{\begin{cases}a>0\\a>\frac{1}{4}\end{cases}\Rightarrow}a>\frac{1}{4}}\)
Vậy để \(\sqrt{A}>A\)thì \(a>\frac{1}{4};a\ne1\)
c,\(\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}.\sqrt{1-a}}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}-1}{a}\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(1+a\right)-\left(1-a\right)}.\frac{\left(\sqrt{1-a^2}-1\right)}{a}=-1\)
M chỉ làm tiếp thôi nha, ko chép lại đề với đk đâu
a,
\(=\frac{a+2\sqrt{ab}+b-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\)\(\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\sqrt{a}+\sqrt{b}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=0\)
b,
\(=\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}+1\right)\)
\(=\left(a-b\right)^2\left(\frac{a+b}{a-b}-1\right)\)
\(=\left(a-b\right)^2\cdot\frac{a+b-a+b}{a-b}\)
\(=\left(a-b\right)2b=2ab-2b^2\)
\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)
Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)
a) \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(ĐK:a>0;a\ne1;a\ne4\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Để \(A>\frac{1}{6}\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Leftrightarrow\)\(\frac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}>0\)
\(\Leftrightarrow\)\(\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
\(\Leftrightarrow\sqrt{a}-4>0\Leftrightarrow a>16\left(tm\right)\)
Vậy a>16 thì \(A>\frac{1}{6}\)
ĐKXĐ : \(a>0,a\ne4,a\ne1\)
a) \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) \(A>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow-\frac{2}{3\sqrt{a}}+\frac{1}{3}>\frac{1}{6}\Rightarrow\frac{2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{1}{\sqrt{a}}>\frac{1}{4}\Rightarrow a< 16\)
Kết hợp với điều kiện xác định.