\(Q=\left(\frac{3}{\sqrt{x}+1}+\frac{5}{\sqrt{x}-1}\right)\div\frac{4\sqrt{x}+1}{x+2\sqrt{x}+1}=\frac{3\sqrt{x}-3+5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}+1}=\frac{2\left(4\sqrt{x}+1\right)}{\sqrt{x}-1}\times\frac{\sqrt{x}+1}{4\sqrt{x}+1}=\frac{2\sqrt{x}+2}{\sqrt{x}-1}\)
Với \(x=3+2\sqrt{2}\Leftrightarrow x=\left(\sqrt{2}+1\right)^2\)
\(\frac{2\sqrt{x}+2}{\sqrt{x}-1}=\frac{2\sqrt{\left(\sqrt{2}+1\right)^2}+2}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}=\frac{2\left(\sqrt{2}+1\right)+2}{\sqrt{2}+1-1}=2+2\sqrt{2}\)
a. Q = \(\left(\frac{3}{\sqrt{x}+1}+\frac{5}{\sqrt{x}-1}\right):\frac{4\sqrt{x}+ 1}{x+2\sqrt{x}+1}\)
\(=\left[\frac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{5\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{4\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}\)
\(=\frac{3\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}+1}\)
\(=\frac{3\sqrt{x}-3+5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}+1}\\ =\frac{8\sqrt{x}+2}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{4\sqrt{x}+1}\\ =\frac{2\left(4\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{4\sqrt{x}+1}\\ =\frac{2\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
b. Với x \(=3+2\sqrt{2}\) (thỏa mãn \(x>0,x\ne1\))
Thay x \(=3+2\sqrt{2}\) vào giá trị biểu thức Q đã rút gọn ta có:
Q \(=\frac{2\left(\sqrt{3+2\sqrt{2}}+1\right)}{\sqrt{3+2\sqrt{2}-1}}=\frac{2\left[\left(\sqrt{2}+1\right)^2+1\right]}{\left(\sqrt{2}+1\right)^2-1}=\frac{2.\left|\sqrt{2}+1\right|+1}{\left|\sqrt{2}+1\right|-1}\\ =\frac{2.\left(\sqrt{2}+1\right)+1}{\sqrt{2}+1-1}=\frac{2\sqrt{2}+3}{\sqrt{2}}=\frac{\left(2\sqrt{2}+3\right)\sqrt{2}}{2}=\frac{4+3\sqrt{2}}{2}\)
Cho mình hỏi đâu là câu trả lời đúng vậy ?