Question

\(F=\dfrac{3}{\sqrt{x-3}-\sqrt{x}}+\dfrac{3}{\sqrt{x-3}+\sqrt{x}}+\dfrac{x\sqrt{x}+x}{\sqrt{x}+1}\)

a)Tìm x sao cho F=6

b)Tìm x sao cho F>2

c)So sánh F với 1,5

Answer 1

điều kiện \(x\ge3\)

\(F=\dfrac{3}{\sqrt{x-3}-\sqrt{x}}+\dfrac{3}{\sqrt{x-3}+\sqrt{x}}+\dfrac{x\sqrt{x}+x}{\sqrt{x}+1}\)

\(F=\dfrac{3\left(\sqrt{x-3}+\sqrt{x}\right)+3\left(\sqrt{x-3}-\sqrt{x}\right)}{\left(\sqrt{x-3}-\sqrt{x}\right)\left(\sqrt{x-3}+\sqrt{x}\right)}+\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(F=\dfrac{3\sqrt{x-3}+3\sqrt{x}+3\sqrt{x-3}-3\sqrt{x}}{\left(\sqrt{x-3}\right)^2-\left(\sqrt{x}\right)^2}+x\)

\(F=\dfrac{6\sqrt{x-3}}{x-3-x}+x=\dfrac{6\sqrt{x-3}}{-3}+x=-2\sqrt{x-3}+x\)