Question

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Answer 1

a.

\(x=9\Rightarrow A=\dfrac{9+8}{\sqrt{9}-1}=\dfrac{17}{2}\)

b.

\(B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{8\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+5\left(\sqrt{x}-1\right)-\left(8\sqrt{x}-6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

c.

\(P=AB=\dfrac{x+8}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{x-1+9}{\sqrt{x}+1}\)

\(P=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+9}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}\)

\(P=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\)

\(P\ge2\sqrt{\dfrac{9\left(\sqrt{x}+1\right)}{\sqrt{x}+1}}-2=4\)

\(P_{min}=4\) khi \(\sqrt{x}+1=3\Rightarrow x=4\)

Answer 2

1: Thay x=9 vào A, ta được:

\(A=\dfrac{9+8}{\sqrt{9}-1}=\dfrac{17}{3-1}=\dfrac{17}{2}\)

2: \(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{5}{\sqrt{x}+1}-\dfrac{8\sqrt{x}-6}{x-1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{5}{\sqrt{x}+1}-\dfrac{8\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+5\left(\sqrt{x}-1\right)-8\sqrt{x}+6}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+5\sqrt{x}-5-8\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

3: \(P=A\cdot B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\cdot\dfrac{x+8}{\sqrt{x}-1}=\dfrac{x+8}{\sqrt{x}+1}\)

\(=\dfrac{x-1+9}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{9}{\sqrt{x}+1}\)

\(=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2\)

\(\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{9}{\sqrt{x}+1}}=6\)

=>\(P=\sqrt{x}+1+\dfrac{9}{\sqrt{x}+1}-2>=6-2=4\)

Dấu '=' xảy ra khi \(\sqrt{x}+1=3\)

=>x=4