Câu1: Rút gọn
\(a,x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\\
b,\sqrt{m^2-6m+9-2m}\left(x3\right)\\
c,1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\\
d,\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
Câu 2: So sánh
\(a,3và\sqrt{5}\\
\\
\\
b,2\sqrt{2}và3\sqrt{2}\\
\\
\\
c,-4\sqrt{5}và-6\sqrt{6}\\
\\
\\
d,2\sqrt{3}-5và\sqrt{3}-4\\
\\
\\e,A=\sqrt{2006}-\sqrt{2005}và\\
B=\sqrt{2005}-\sqrt{2004}\)
Câu 3: Rút gọn
\(a,\sqrt{16-2\sqrt{55}}\\ \\ \\ \\ \\ \\ \\ \\ \\...
Đọc tiếp
Câu1: Rút gọn
\(a,x+\sqrt{\left(x+2\right)^2}\cdot\left(x-2\right)\\
b,\sqrt{m^2-6m+9-2m}\left(x>3\right)\\
c,1+\sqrt{\frac{\left(x-1\right)^2}{x-1}}\\
d,\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)
Câu 2: So sánh
\(a,3và\sqrt{5}\\
\\
\\
b,2\sqrt{2}và3\sqrt{2}\\
\\
\\
c,-4\sqrt{5}và-6\sqrt{6}\\
\\
\\
d,2\sqrt{3}-5và\sqrt{3}-4\\
\\
\\e,A=\sqrt{2006}-\sqrt{2005}và\\
B=\sqrt{2005}-\sqrt{2004}\)
Câu 3: Rút gọn
\(a,\sqrt{16-2\sqrt{55}}\\ \\ \\ \\ \\ \\ \\ \\ \\ b,\sqrt{14-6\sqrt{5}}\\
\\
\\
\\
\\
\\
\\
\\
\\
c,\sqrt{36+12\sqrt{5}}\\
\\
\\
\\
\\
\\
\\
\\
\\
d,\sqrt{29+12\sqrt{5}}\)
Câu4: Tìm đkxđ
\(a,\sqrt{x^2-9}\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
b,\sqrt{x^2-3x+2}\)
\(c,\frac{\sqrt{x+3}}{\sqrt{5-x}}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ d,\sqrt{\frac{x+3}{5-x}}\)
2) \(-x^2+4x-2\)
\(=-\left(x^2-4x+2\right)\)
\(=-\left(x^2-4x+4-2\right)\)
\(=-\left(x-2\right)^2+2\)
Ta có: \(-\left(x-2\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-2\right)^2+2\le2\forall x\)
Dấu "=" xảy ra:
\(\Leftrightarrow-\left(x-2\right)^2+2=2\Leftrightarrow x=2\)
Vậy: GTLN của bt là 2 tại x=2
b) \(\sqrt{2x^2-3}\) (ĐK: \(x\ge\sqrt{\dfrac{3}{2}}\))
Mà: \(\sqrt{2x^2-3}\ge0\forall x\)
Dấu "=" xảy ra:
\(\sqrt{2x^2-3}=0\Leftrightarrow x=\sqrt{\dfrac{3}{2}}=\dfrac{3\sqrt{2}}{2}\)
Vậy GTNN của bt là 0 tại \(x=\dfrac{3\sqrt{2}}{2}\)
...
1:
b: \(4\sqrt{5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{75}\)
=>\(4\sqrt{5}>5\sqrt{3}\)
=>\(\sqrt{4\sqrt{5}}>\sqrt{5\sqrt{3}}\)
c: \(3-2\sqrt{5}-1+\sqrt{5}=2-\sqrt{5}< 0\)
=>\(3-2\sqrt{5}< 1-\sqrt{5}\)
d: \(\sqrt{2006}-\sqrt{2005}=\dfrac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2005}-\sqrt{2004}=\dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)
\(\sqrt{2006}+\sqrt{2005}>\sqrt{2005}+\sqrt{2004}\)
=>\(\dfrac{1}{\sqrt{2006}+\sqrt{2005}}< \dfrac{1}{\sqrt{2005}+\sqrt{2004}}\)
=>\(\sqrt{2006}-\sqrt{2005}< \sqrt{2005}-\sqrt{2004}\)
e: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)