\(\frac{\sqrt{a+3+4\sqrt{a-1}}}{\sqrt{a-1}+2}\)
- Chung minh bieu thuc tren bang 1 voi a >= 1
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24 tháng 11 2019
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
24 tháng 11 2019
a/ Ta có: A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)=\left(\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}+1\right):\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\)
\(=\left(\sqrt{x}+1\right):\left(\sqrt{x}\right)=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b/ Ta có :\(x=7+4\sqrt{3}=3+4\sqrt{3}+4=\left(\sqrt{3}+2\right)^2
\)
\(\Rightarrow\sqrt{x}=|\sqrt{3}+2|=\sqrt{3}+2\)
Thay x vào A ta có:
A\(=\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}+2+1}{\sqrt{3}+2}=\frac{\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{3-\sqrt{3}}{1}=3-\sqrt{3}\)
7 tháng 8 2020
Bài 1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
a) Ta có: \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)
\(=\frac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
Để A>1 thì A-1>0
\(\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-2}-1>0\)
\(\Leftrightarrow\frac{\sqrt{x}-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\frac{2}{\sqrt{x}-2}>0\)
mà 2>0
nên \(\sqrt{x}-2>0\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4(nhận)
Vậy: Khi x>4 thì A>1
RR
13 tháng 5 2018
a) Với x = 25 thì \(N=\frac{\sqrt{25}+1}{\sqrt{25}}=\frac{6}{5}\)
b) Ta có \(M=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2.\left(\sqrt{x}-1\right)}\)
\(M=\frac{2\sqrt{x}}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\)
Suy ra \(S=M.N=\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(\frac{\sqrt{a+3+4\sqrt{a-1}}}{\sqrt{a-1}+2}\)
\(\frac{\sqrt{\left(\sqrt{a-1}+2\right)^2}}{\sqrt{a-1}+2}=\frac{\sqrt{a-1}+2}{\sqrt{a-1}+2}=1\)
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