a) \(B=\dfrac{x\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{x-1}\Rightarrow B=\dfrac{x\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Rightarrow B=\dfrac{x\sqrt{x}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Rightarrow B=\dfrac{\left(\sqrt{x}-1\right)\left(x\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Rightarrow B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Rightarrow B=x-\sqrt{x}+1\)b) ta có x=\(7-4\sqrt{3}\Rightarrow B=7-4\sqrt{3}-\sqrt{7-4\sqrt{3}}+1=8-4\sqrt{3}-\sqrt{\left(2-\sqrt{3}\right)^2}=8-4\sqrt{3}-2+\sqrt{3}=6-3\sqrt{3}\)c) ta có \(B=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy GTNN của B là \(\dfrac{3}{4}\)
Đẳng thức xảy ra khi \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\Rightarrow\sqrt{x}-\dfrac{1}{2}=0\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
