So sánh các số sau :
a)\(\sqrt{7}-\sqrt{2}\)và 1
b)\(\sqrt{8}+\sqrt{5}\)và \(\sqrt{7}+\sqrt{6}\)
c) \(\sqrt{2005}+\sqrt{2007}\)và \(\sqrt{2006}\)
Bài 2 so sánh giải từng bước giúp mình nha...
a\(\sqrt{7}-\sqrt{2}và1\)
b \(\sqrt{8}+\sqrt{5}và\sqrt{7}+\sqrt{6}\)
c \(\sqrt{2005}+\sqrt{2007}và2\sqrt{2006}\)
d \(\sqrt{16+9}và\sqrt{16}+\sqrt{9}\)................các bạn ơi giải nhanh giúp mình với hepl me.....
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: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
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=\(\dfrac{\sqrt{99}-\sqrt{3}}{2}\)
\(\sqrt{2006}-\sqrt{2005}và\sqrt{2008}-\sqrt{2007}\)
So sánh
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Easy
Ta có:
\(\sqrt{2006}-\sqrt{2005}=\frac{2006-2005}{\sqrt{2006}+\sqrt{2005}}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
Tương tự cũng có: \(\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
Dễ thấy: \(\sqrt{2005}+\sqrt{2006}< \sqrt{2007}+\sqrt{2008}\)
\(\Rightarrow\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2008}}\)
rút gọn
1/\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
2.\(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
so sánh 1/ \(\sqrt{7}-\sqrt{6}với\sqrt{3}-\sqrt{2}\)
2/\(\sqrt{2005}+\sqrt{2007}với2\sqrt{2006}\)
3/\(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}với\sqrt{2}+1\)
giải giúp mình đi mai là mình đi học rồi
Ta có:
bla bla ........
vậy đáp số là... quên mất rồi
So sánh \(\sqrt{2005}+\sqrt{2007}\) và \(2\sqrt{2006}\)
Ta có : \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Mà : \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}-\sqrt{2006}}\)
Nến : \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
So sánh \(\sqrt{2005}+\sqrt{2007}\) và \(2\sqrt{2006}\)
\(\left(\sqrt{2005}+\sqrt{2007}\right)^2=4012+2\sqrt{2005.2007}\)
\(=4012+2\sqrt{\left(2016-1\right)\left(2016+1\right)}=4012+2\sqrt{2016^2-1}\)
\(\left(2\sqrt{2006}\right)^2=4012+4012=4012+2\sqrt{2016^2}\)
=>\(\left(\sqrt{2015}+\sqrt{2017}\right)^2< \left(2\sqrt{2016}\right)^2\Rightarrow\sqrt{2015}+\sqrt{2017}< 2\sqrt{2016}\)
Ta có: \(\sqrt{2006}-\sqrt{2005}=\frac{1}{\sqrt{2006}+\sqrt{2005}}\)
\(\sqrt{2007}-\sqrt{2006}=\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Mà: \(\frac{1}{\sqrt{2006}+\sqrt{2005}}>\frac{1}{\sqrt{2007}+\sqrt{2006}}\)
Nên: \(\sqrt{2006}-\sqrt{2005}>\sqrt{2007}-\sqrt{2006}\)
=>\(\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)
so sánh
\(\sqrt{2}+\sqrt{3}\) và 2
\(\sqrt{8}+\sqrt{5}\) và \(\sqrt{7}-\sqrt{6}\)
\(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}>2^2=4\left(5>4\right)\\ \Leftrightarrow\sqrt{2}+\sqrt{3}>2\)
\(\left(\sqrt{8}+\sqrt{5}\right)^2=13+2\sqrt{40};\left(\sqrt{7}-\sqrt{6}\right)^2=13-2\sqrt{42}\\ 2\sqrt{40}>0>-2\sqrt{42}\\ \Leftrightarrow13+2\sqrt{40}>13-2\sqrt{42}\\ \Leftrightarrow\left(\sqrt{8}+\sqrt{5}\right)^2>\left(\sqrt{7}-\sqrt{6}\right)^2\\ \Leftrightarrow\sqrt{8}+\sqrt{5}>\sqrt{7}-\sqrt{6}\)
\(\sqrt{2}\) + \(\sqrt{3}\) > 2
So sánh các số:
a)\(\sqrt{7}\) - \(\sqrt{2}\) và 1;
b) \(\sqrt{8}\) + \(\sqrt{5}\) và \(\sqrt{7}\) + \(\sqrt{6}\);
c) \(\sqrt{2005}\) + \(\sqrt{2007}\) và 2\(\sqrt{2006}\).
\(\text{a) Ta có }:\left(\sqrt{7}-\sqrt{2}\right)^2=7-\sqrt{14}+2=9-\sqrt{14}\\ 1^2=1=9-8=9-\sqrt{64}\\ Do\text{ }\sqrt{14}< \sqrt{64}\Rightarrow9-\sqrt{14}>9-\sqrt{64}\\ \Rightarrow\left(\sqrt{7}-\sqrt{2}\right)^2>1^2\\ \Rightarrow\sqrt{7}-\sqrt{2}>1\)
\(\text{b) Ta có: }\left(\sqrt{8}+\sqrt{5}\right)^2=8+\sqrt{160}+5=13+\sqrt{160}\\ \left(\sqrt{7}+\sqrt{6}\right)^2=7+\sqrt{168}+6=13+\sqrt{168}\\ \text{Do }\sqrt{160}< \sqrt{168}\Rightarrow13+\sqrt{160}< 13+\sqrt{168}\\ \Rightarrow\left(\sqrt{8}+\sqrt{5}\right)^2< \left(\sqrt{7}+\sqrt{6}\right)^2\\ \Rightarrow\sqrt{8}+\sqrt{5}< \sqrt{7}+\sqrt{6}\)
\(\text{c) Ta có }:\left(\sqrt{2005}+\sqrt{2007}\right)^2\\ =2005+2\sqrt{2005\cdot2007}+2007\\ =4012+2\sqrt{2005\cdot2007}\\ \left(2\sqrt{2006}\right)^2=4\cdot2006=4012+2\cdot2006\)
\(\text{Lại có }:\sqrt{2005\cdot2007}=\sqrt{\left(2006-1\right)\left(2006+1\right)}=\sqrt{2006^2-1}\\ Do\text{ }\sqrt{2006^2-1}< \sqrt{2006^2}\\ \Rightarrow\sqrt{2005\cdot2007}< 2006\\ \Rightarrow2\sqrt{2005\cdot2007}< 2\cdot2006\\ \Rightarrow4012+2\sqrt{2005\cdot2007}< 4012+2\cdot2006\\ \Rightarrow\left(\sqrt{2005}+\sqrt{2007}\right)^2< \left(2\sqrt{2006}\right)^2\\ \Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)