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\(1,P=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\frac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
a/ \(A=\left(\frac{2\sqrt{x}+x}{\sqrt{x}^3-1}-\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\left[\frac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)
\(=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}-1}.\frac{1}{\sqrt{x}+2}=\frac{1}{\sqrt{x}+2}\)
b/ Thay \(x=4+2\sqrt{3}\) vào A ta được:
\(A=\frac{1}{\sqrt{4+2\sqrt{3}}+2}=\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}+2}=\frac{1}{\sqrt{3}+3}\)
\(\Rightarrow\sqrt{A}=\frac{1}{\sqrt{\sqrt{3}+3}}\)
Câu 1 bạn ghi lại đề, tử số cuối cùng bị thiếu
2/ \(a^2-4a+1=0\Rightarrow a^2+1=4a\Rightarrow a+\frac{1}{a}=4\)
\(\Rightarrow a^2+\frac{1}{a^2}+2=16\Rightarrow a^2+\frac{1}{a^2}=14\)
\(P=\frac{a^4+a^2+1}{a^2}=a^2+\frac{1}{a^2}+1=14+1=15\)
Lời giải:
a)
Ta có: \(\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}=\frac{\sqrt{3}-2+\sqrt{3}+2}{(\sqrt{3}+2)(\sqrt{3}-2)}=\frac{2\sqrt{3}}{3-4}=-2\sqrt{3}\)
Để \(B=\frac{1}{\sqrt{3}+2}+\frac{1}{\sqrt{3}-2}\Leftrightarrow \frac{2}{\sqrt{x}-2}=-2\sqrt{3}\)
\(\Leftrightarrow \frac{1}{\sqrt{x}-2}=-\sqrt{3}\)
\(\Leftrightarrow\sqrt{x}-2=\frac{-1}{\sqrt{3}}\)
\(\Leftrightarrow \sqrt{x}=2-\frac{1}{\sqrt{3}}\Rightarrow x=(2-\frac{1}{\sqrt{3}})^2=\frac{13-4\sqrt{3}}{3}\)
b)
ĐK: \(x\geq 0; x\neq 4\)
\(A=\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}=\frac{2\sqrt{x}+2}{x-4}\)
\(P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\frac{2(\sqrt{x}+1)}{x-4}=\frac{2(x-4)}{2(\sqrt{x}-2)(\sqrt{x}+1)}\)
\(=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+1)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c) Thêm ĐK: \(x\geq 1\)
Từ biểu thức P vừa tìm được:
\(P(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \frac{\sqrt{x}+2}{\sqrt{x}+1}.(\sqrt{x}+1)-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow \sqrt{x}+2-\sqrt{x}+2\sqrt{x-1}=2x-2\sqrt{2x}+4\)
\(\Leftrightarrow 2\sqrt{x-1}=2x-2\sqrt{2x}+2\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2=0\)
Vì \((\sqrt{x-1}-1)^2, (\sqrt{x}-\sqrt{2})^2\geq 0, \forall x\in \text{ĐKXĐ}\)
\(\Rightarrow (\sqrt{x-1}-1)^2+(\sqrt{x}-\sqrt{2})^2\geq 0\). Dấu bằng xảy ra khi :
\(\left\{\begin{matrix} \sqrt{x-1}-1=0\\ \sqrt{x}-\sqrt{2}=0\end{matrix}\right.\Leftrightarrow x=2\) (thỏa mãn)
Vậy..........
