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 \(\sqrt{a}+2\ge2\left(\forall a\right)\)

\(\Rightarrow3\sqrt{a}-1=0\Leftrightarrow3\sqrt{a}=1\)

\(\Leftrightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)

\(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)

\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a-\sqrt{a}\right)\left(a\sqrt{a}+1\right)}{\left(a-\sqrt{a}\right)\left(a+\sqrt{a}\right)}\)

\(=\frac{a^2\cdot\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}+a-a^2-\sqrt{a}\right)}{a^2-a}\)

\(=\frac{2a^2-2a}{a^2-a}\)

\(=2\)( 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

\(2+\frac{2\left(a+1\right)}{\sqrt{a}}=7\Leftrightarrow2a+2=5\sqrt{a}\)

\(\Leftrightarrow2a-5\sqrt{a}+2=0\)

\(\Leftrightarrow\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\Rightarrow\orbr{\begin{cases}a=\frac{1}{4}\\a=4\end{cases}}\)

\(2+\frac{2\left(a+1\right)}{\sqrt{a}}>6\)\(\Rightarrow2a+2>4\sqrt{a}\Rightarrow2\left(a+1-2\sqrt{a}\right)>0\)

\(\Leftrightarrow\left(a+1-2\sqrt{a}\right)>0\Leftrightarrow\left(\sqrt{a}-1\right)^2>0\)

\(\Leftrightarrow a\ne1;a\ge0\)

a: \(A=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)

\(=2+\dfrac{2a+2}{\sqrt{a}}=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

b: Để A=7 thì \(2a-5\sqrt{a}+2=0\)

\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)

=>a=4 hoặc a=1/4

Em kiểm tra lại đề bài nhé!