a: \(B=\dfrac{2x+\sqrt{x}-1}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\)
\(=\left(2x+\sqrt{x}-1\right)\left(\dfrac{-1}{x-1}+\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\cdot\left(\dfrac{-x+\sqrt{x}-1+x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=-\dfrac{\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}\)
\(A=\dfrac{-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\dfrac{-x+\sqrt{x}-1}{\sqrt{x}}\)
b: Khi \(x=17-12\sqrt{2}=\left(3-2\sqrt{2}\right)^2\) thì
\(A=\dfrac{-17+12\sqrt{2}+3-2\sqrt{2}-1}{3-2\sqrt{2}}=-5\)
c: \(A=\dfrac{-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{-\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{\sqrt{x}}< 0\)
=>căn A không tồn tại
