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\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
Trả lời:
b, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x>0\right)\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1+2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)
c, \(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}>\frac{3}{2}\) \(\left(ĐK:x>0\right)\)
\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}>\frac{3}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)
\(\Leftrightarrow\frac{2\left(\sqrt{x}+1\right)-3\sqrt{x}}{2\sqrt{x}}>0\)
\(\Rightarrow2\sqrt{x}+1-3\sqrt{x}>0\Leftrightarrow1-\sqrt{x}>0\)
\(\Leftrightarrow-\sqrt{x}>-1\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)
Vậy \(0< x< 1\) là giá trị cần tìm.
1. \(VT=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}-2+\sqrt{3}=VP\)
Bài 1.
Ta có : \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3+4\sqrt{3}+4}-\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|\sqrt{3}-2\right|\)
\(=\sqrt{3}+2-\left(2-\sqrt{3}\right)\)
\(=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\left(đpcm\right)\)
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:
\(A=\frac{1}{\sqrt{4+2\sqrt{3}}+2}=\frac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}+2}=\frac{1}{\sqrt{3}+3}\)
\(\Rightarrow\sqrt{A}=\frac{1}{\sqrt{\sqrt{3}+3}}\)
\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)\) phải bằng \(x-4\) chứ!!
\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)=\left(\sqrt{x}\right)^2-2\sqrt{x}+2\sqrt{x}+2^2=x-4\)