Tính: a, \(\sqrt{2006+2\sqrt{2005}}-\sqrt{2006-2\sqrt{2005}}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
Rút gọn:
a) \(A=\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+... +\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
b) \(B=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2006\sqrt{2005}+2005\sqrt{2006}}+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\)
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
Tiếp phần b ( do máy lag) :3
Cộng 2 vế với nhau, ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}\\ =1-\dfrac{1}{\sqrt{2007}}\)
a) A=\(\dfrac{1}{\sqrt{3}+\sqrt{5}}\)+\(\dfrac{1}{\sqrt{5}+\sqrt{7}}\)+\(\dfrac{1}{\sqrt{7}+\sqrt{9}}\)+...+\(\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)+\(\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)+\(\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\left(\sqrt{9}-\sqrt{7}\right)}\)+...+\(\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}}{2}\)
=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
vậy A=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
1/ So sánh
a) 3 - 2\(\sqrt{3}\) và 2\(\sqrt{6}\) - 5
b) \(\sqrt{4\sqrt{5}}\) và \(\sqrt{5\sqrt{3}}\)
c) 3 - 2\(\sqrt{5}\) và 1 - \(\sqrt{5}\)
d) \(\sqrt{2006}\) - \(\sqrt{2005}\) và \(\sqrt{2005}\) - \(\sqrt{2004}\)
e) \(\sqrt{2003}\) + \(\sqrt{2005}\) và \(2\sqrt{2004}\)
2/ Tìm giá trị nhỏ nhất hoặc giá trị lớn nhất
a) -x² + 4x - 2
b) \(\sqrt{2x^2\:+\:3}\)
c) 2x - \(\sqrt{1x}\)
d) -3 + \(\sqrt{2x^2\:+\:49}\)
e) \(\sqrt{9x^2\:-\:4x\:+\:65}\)
f) -5 + \(\sqrt{4\:-\:9x^2\:+\:6x}\)
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: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}+\dfrac{2005}{2006}}\)
Sửa đề:
\(VP=\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)
Ta có: \(2005^2+1=\left(2005+1\right)^2-2.2005.1=2006^2-2.2005\)
\(\Rightarrow VP=\sqrt{2006^2-2.2005+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)
\(=\sqrt{\left(2006-\dfrac{2005}{2006}\right)^2}+\dfrac{2005}{2006}\)
\(=2006-\dfrac{2005}{2006}+\dfrac{2005}{2006}=2006\)
Phương trình đã cho tương đương
\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=2006\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2006\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
Đến đây thì tự xét trường hợp và giải tìm nghiệm, bài này không cần điều kiện nhé
tính
a/ (\(\sqrt{2006}\)- \(\sqrt{2005}\)).(\(\sqrt{2006}\)+ \(\sqrt{2005}\))
b/ (\(\frac{1}{7-4\sqrt{3}}\)+ \(\frac{2}{7+4\sqrt{3}}\)).(21+4\(\sqrt{3}\))
c/ \(3\sqrt{\frac{9}{8}}\)- \(\sqrt{\frac{49}{2}}\)+ \(\sqrt{\frac{25}{18}}\)
d/ ( \(5\sqrt{28}-2\sqrt{63}+3\sqrt{112}\)) : \(\sqrt{7}\)
e/ \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
f/ (\(\frac{4\sqrt{21}-4\sqrt{15}-\sqrt{14}+\sqrt{10}}{4\sqrt{6}-2+4\sqrt{15}-\sqrt{10}}\)
tính
a/ (\(\sqrt{2006}\)- \(\sqrt{2005}\)).(\(\sqrt{2006}\)+ \(\sqrt{2005}\))
b/ (\(\frac{1}{7-4\sqrt{3}}\)+ \(\frac{2}{7+4\sqrt{3}}\)).(21+4\(\sqrt{3}\))
c/ \(3\sqrt{\frac{9}{8}}\)- \(\sqrt{\frac{49}{2}}\)+ \(\sqrt{\frac{25}{18}}\)
d/ ( \(5\sqrt{28}-2\sqrt{63}+3\sqrt{112}\)) : \(\sqrt{7}\)
e/ \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)
f/ (\(\frac{4\sqrt{21}-4\sqrt{15}-\sqrt{14}+\sqrt{10}}{4\sqrt{6}-2+4\sqrt{15}-\sqrt{10}}\)
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)
giai phuong trinh \(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}+\dfrac{2005}{2006}\)
\(\sqrt{1+2005^2+\dfrac{2005^2}{2006^2}}=\dfrac{1}{2006}\sqrt{2006^2+2005^2+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2006-2005\right)^2+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{1+2.2005.2006+\left(2005.2006\right)^2}\)
\(=\dfrac{1}{2006}\sqrt{\left(2005.2006+1\right)^2}=\dfrac{2005.2006+1}{2006}=2005+\dfrac{1}{2006}\)
Phương trình tương đương:
\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x-2\right)^2}=2005+\dfrac{1}{2006}+\dfrac{2005}{2006}\)
\(\Leftrightarrow\left|x-1\right|+\left|x-2\right|=2006\)
TH1: \(x\ge2\): \(x-1+x-2=2006\Rightarrow2x=2009\Rightarrow x=\dfrac{2009}{2}\)
TH2: \(x\le1\) : \(1-x+2-x=2006\Rightarrow-2x=2003\Rightarrow x=\dfrac{-2003}{2}\)
TH3: \(1< x< 2:\) \(x-1+2-x=2006\Rightarrow3=2006\) (vô nghiệm)
Vậy \(\left[{}\begin{matrix}x=\dfrac{2009}{2}\\x=\dfrac{-2003}{2}\end{matrix}\right.\)
so sánh
a\(\sqrt{1999}+\sqrt{2001}\) Và \(2\sqrt{2000}\)
b \(\frac{2006}{\sqrt{2005}}+\frac{2005}{\sqrt{2006}}\)Và \(\sqrt{2005+\sqrt{2006}}\)
Chứng tỏ rằng
\(\frac{2}{3\left(1+\sqrt{2}\right)}+\frac{2}{5\left(\sqrt{2}+\sqrt{3}\right)}+\frac{2}{7\left(\sqrt{3}+\sqrt{4}\right)}+...+\frac{2}{4011\left(\sqrt{2005}+\sqrt{2006}\right)}<1-\frac{1}{\sqrt{2006}}\)