giải dùm mình câu 1 với ạ
1) \(Q=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]\cdot\dfrac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{.\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x-2\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(x-2\right)+\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-2\sqrt{x}+x\sqrt{x}+x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}+x-\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\)
b) \(x=2\left(3-\sqrt{5}\right)=6-2\sqrt{5}=\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2=\left(\sqrt{5}-1\right)^2\)
Thay x vào Q ta có:
\(Q=\dfrac{6-2\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}{6-2\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}+1}\)
\(=\dfrac{6-2\sqrt{5}+\sqrt{5}-1}{6-2\sqrt{5}+\sqrt{5}-1+1}=\dfrac{5-\sqrt{5}}{6-\sqrt{5}}=\dfrac{25-\sqrt{5}}{31}\)
c) Ta có:
\(Q=\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}=\dfrac{x+\sqrt{x}+1-1}{x+\sqrt{x}+1}=1-\dfrac{1}{x+\sqrt{x}+1}\)
Theo đkxđ: \(x\ge0\Rightarrow x+\sqrt{x}\ge0\Rightarrow x+\sqrt{x}+1\ge1\)
\(\Rightarrow\dfrac{1}{x+\sqrt{x}+1}\le1\Rightarrow1-\dfrac{1}{x+\sqrt{x}+1}\ge1-1=0\)
Dấu "=" xảy ra: `x=0`
Vậy: \(Q_{min}=0\Leftrightarrow x=0\)