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a) với x=25 thỏa mãn điều kiện xác định
Thay x=25 vào biều thức A ta có:
A=\(\dfrac{\sqrt{25}-3}{\sqrt{25}+1}=\dfrac{5-3}{5+1}=\dfrac{2}{6}=\dfrac{1}{3}\)
Vậy giá trị của biểu thức A tại x=25 là\(\dfrac{1}{3}\)
b) Ta có B=\(\left(\dfrac{x}{x-4}-\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
=\(\left(\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\times\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
=\(\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
=\(\dfrac{x-\sqrt{x}-2}{x-2\sqrt{x}}\)
=\(\dfrac{x-2\sqrt{x}+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\text{}\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\text{}\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\text{}\dfrac{\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
Vậy B=\(\text{}\dfrac{\left(\sqrt{x}+1\right)}{\sqrt{x}}\) với x≠4 và x≥0
c)