1) Biểu thức này là P hả?
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
P = \(\dfrac{\sqrt{a^3}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\sqrt{a^3}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\dfrac{a-1}{\sqrt{a}}\right).\left(\dfrac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{a-1}\right)\)
= \(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\sqrt{a}}\)= \(\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)+2a+2}{\sqrt{a}}\)
= \(\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1+2a+2}{\sqrt{a}}\)
= \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
2) Để P = 7 với a ∈ ĐKXĐ
⇒ \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\) = 7
⇔ 2a + 2√a+2 = 7√a
⇔ 2a - 5√a + 2 = 0
⇔ \(\left[{}\begin{matrix}a=2\\a=\dfrac{1}{2}\end{matrix}\right.\)( thoả mãn ĐKXĐ)
Vậy...
3) Để P > 6 với a ∈ ĐKXĐ
⇒ \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\) >6
⇔ \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\) - 6 > 0
⇔ \(\dfrac{2a+2\sqrt{a}-6\sqrt{a}+2}{\sqrt{a}}>0\)
Mà √a > 0 với ∀a ∈ ĐKXĐ
⇒ 2a - 4√a + 2 >0
⇔ 2(√a - 1)2 > 0
Do 2(√a - 1)2 ≥ 0 với ∀a ∈ ĐKXĐ
Nên để 2(√a - 1)2 > 0 ⇔ 2(√a - 1)2 ≠ 0
⇔ a ≠ 1
Đối chiếu ĐKXĐ ta được: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
Vậy để P > 6 thì a ∈ ĐKXĐ
ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
1) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\left(\dfrac{a}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\left(\dfrac{a+2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\dfrac{a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{2\sqrt{a}}{\sqrt{a}}+\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
2) Để P=7 thì \(\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\)
\(\Leftrightarrow2a+2\sqrt{a}+2=7\sqrt{a}\)
\(\Leftrightarrow2a+2\sqrt{a}-7\sqrt{a}+2=0\)
\(\Leftrightarrow2a-5\sqrt{a}+2=0\)
\(\Leftrightarrow2a-4\sqrt{a}-\sqrt{a}+2=0\)
\(\Leftrightarrow2\sqrt{a}\left(\sqrt{a}-2\right)-\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{a}-2\right)\left(2\sqrt{a}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-2=0\\2\sqrt{a}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=2\\2\sqrt{a}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\\\sqrt{a}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=4\left(nhận\right)\\a=\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)
Vậy: Để P=7 thì \(a\in\left\{4;\dfrac{1}{4}\right\}\)
