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\(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a-\sqrt{a}\right)\left(a\sqrt{a}+1\right)}{\left(a-\sqrt{a}\right)\left(a+\sqrt{a}\right)}\)
\(=\frac{a^2\cdot\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}+a-a^2-\sqrt{a}\right)}{a^2-a}\)
\(=\frac{2a^2-2a}{a^2-a}\)
\(=2\)( 1 )
\(\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\left(\frac{\sqrt{a}}{1}-\frac{1}{\sqrt{a}}\right)\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\right)\)
\(=\frac{a-1}{\sqrt{a}}\cdot\frac{2\left(a+1\right)}{a-1}\)
\(=\frac{2\left(a+1\right)}{\sqrt{a}}\) ( 2 )
Cộng ( 1 ) và ( 2 ) lại thì ta được biểu thức ban đầu:
\(2+\frac{2\left(a+1\right)}{\sqrt{a}}\)
Câu b,c em chịu:((
P/S:e ko bt đúng hay sai đâu ạ
Mk giải nốt phần còn lại nha
sai thì thông cảm
\(2+\frac{2\left(a+1\right)}{\sqrt{a}}=7\Leftrightarrow2a+2=5\sqrt{a}\)
\(\Leftrightarrow2a-5\sqrt{a}+2=0\)
\(\Leftrightarrow\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\Rightarrow\orbr{\begin{cases}a=\frac{1}{4}\\a=4\end{cases}}\)
\(2+\frac{2\left(a+1\right)}{\sqrt{a}}>6\)\(\Rightarrow2a+2>4\sqrt{a}\Rightarrow2\left(a+1-2\sqrt{a}\right)>0\)
\(\Leftrightarrow\left(a+1-2\sqrt{a}\right)>0\Leftrightarrow\left(\sqrt{a}-1\right)^2>0\)
\(\Leftrightarrow a\ne1;a\ge0\)
\(đkxđ\Leftrightarrow x\ge0;x\ne1;x\ne4\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(=\left(\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\)\(\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A< \frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>\frac{1}{6}\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}-\frac{1}{6}>0\)
\(\Rightarrow\frac{2\left(\sqrt{a}-2\right)}{6\sqrt{a}}-\frac{\sqrt{a}}{6\sqrt{a}}>0\Rightarrow\frac{\sqrt{a}-4}{6\sqrt{a}}>0\)
Vì \(6\sqrt{a}>0\Rightarrow\sqrt{a}-4>0\Rightarrow\sqrt{a}>4\Rightarrow a>16\)
Vậy \(P>\frac{1}{6}\Leftrightarrow a>16\)
Điều kiện : a> 0 ; a khác 1
\(A=\frac{\left(\sqrt{a}\right)^3-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}\right)^3+1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(A=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\left(\frac{a-1}{\sqrt{a}}\right)\left(\frac{2a+2}{a-1}\right)\)
\(A=\frac{2\sqrt{a}}{\sqrt{a}}+\frac{2\left(a+1\right)}{\sqrt{a}}=2+\frac{2\sqrt{a}\left(a+1\right)}{a}\)
a) đkxđ: \(a>0;a\ne1\)
Ta có:
\(P=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(1-\frac{1}{\sqrt{a}}\right)\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(P=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\frac{\sqrt{a}-1}{\sqrt{a}}.\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(P=\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}+\frac{2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
\(P=\frac{2\sqrt{a}\left(\sqrt{a}+1\right)+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
\(P=\frac{2a+2\sqrt{a}+2a+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
\(P=\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}\)
b) \(P=7\)
\(\Leftrightarrow\frac{4a+2\sqrt{a}+2}{\left(\sqrt{a}+1\right)\sqrt{a}}=7\)
\(\Leftrightarrow4a+2\sqrt{a}+2=7a+7\sqrt{a}\)
\(\Leftrightarrow3a+5\sqrt{a}-2=0\)
\(\Leftrightarrow\left(3a-\sqrt{a}\right)+\left(6\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(3\sqrt{a}-1\right)\sqrt{a}+2\left(3\sqrt{a}-1\right)=0\)
\(\Leftrightarrow\left(3\sqrt{a}-1\right)\left(\sqrt{a}+2\right)=0\)
Mà \(\sqrt{a}+2\ge2\left(\forall a\right)\)
\(\Rightarrow3\sqrt{a}-1=0\Leftrightarrow3\sqrt{a}=1\)
\(\Leftrightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)