Question

Answer 1

a: \(A=\sqrt{7}+2+3-\sqrt{7}=5\)

\(B=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}\cdot\dfrac{x-1}{\sqrt{x}+1}\)

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

Answer 2

\(A=2+\sqrt{7}+\sqrt{7}-3=2\sqrt{7}-1\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x-\sqrt{x}}{x-\sqrt{x}}\cdot\left(\sqrt{x}-1\right)=\sqrt{x}-1\)
Để B≤A thì \(\sqrt{x}-1\le2\sqrt{7}-1\left(x\ge0\right)< =>\sqrt{x}\le2\sqrt{7}-1+1< =>\sqrt{x}\le2\sqrt{7}< =>x\le28\)
Đối chiếu ĐKXĐ 0≤x≥28