Question

Cho A =\(\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\) và B \(\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}+4}{2-\sqrt{x}-x}\) với x ≥0;x ≠1

a) rút gọn B

b) Cho P=A.B. Tìm x ∈ Z⁺ để P min 

Answer 1

a) Với \(x\ge0;x\ne1\):

\(B=\dfrac{2\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}+4}{2-\sqrt{x}-x}\)

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

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

\(=\dfrac{x+2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)

b) \(P=A\cdot B=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\dfrac{\left(\sqrt{x}+2\right)-7}{\sqrt{x}+2}=1-\dfrac{7}{\sqrt{x}+2}\)

Ta thấy: \(\sqrt{x}\ge0;\forall x\ge0\Rightarrow\sqrt{x}+2\ge2;\forall x\ge0\)

\(\Rightarrow\dfrac{7}{\sqrt{x}+2}\le\dfrac{7}{2};\forall x\ge0\)

\(\Rightarrow1-\dfrac{7}{\sqrt{x}+2}\ge1-\dfrac{7}{2}=-\dfrac{5}{2};\forall x\ge0\)

\(\Rightarrow P\ge-\dfrac{5}{2};\forall x\ge0\)

Dấu \("="\) xảy ra khi \(x=0\) (tm ĐKXĐ)

Vậy \(P_{min}=-\dfrac{5}{2}\) tại \(x=0\)

Answer 2

a, đk x >= 0 ; x khác 1

\(\dfrac{2x+4\sqrt{x}-x+1-2\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x+2\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)

b, Ta có \(P=\dfrac{\sqrt{x}-5}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2-7}{\sqrt{x}+2}=1-\dfrac{7}{\sqrt{x}+2}\)

Ta có \(x\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{-7}{\sqrt{x}+2}\ge-\dfrac{7}{2}\Rightarrow P\ge-\dfrac{7}{2}+1=-\dfrac{5}{2}\)

Dấu ''='' xảy ra khi x = 0