So sánh
a)\(4\)và \(1+2\sqrt{2}\)
b)\(4\)và \(2\sqrt{6}-1\)
c)\(\sqrt{0,5}\)và \(\sqrt{3}-2\)
d)\(\sqrt{2015}+\sqrt{2018}\)và \(\sqrt{2016}+\sqrt{2017}\)
So sánh
a.2\(\sqrt{29}\) và 3\(\sqrt{13}\)
b.\(\dfrac{5}{4}\)\(\sqrt{2}\) và \(\dfrac{3}{2}\)\(\sqrt{\dfrac{3}{2}}\)
c.5\(\sqrt{2}\) và 4\(\sqrt{3}\)
d.\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}\) và 6\(\sqrt{\dfrac{1}{37}}\)
a)
Có:
\(2\sqrt{29}=\sqrt{4.29}=\sqrt{116}\\ 3\sqrt{13}=\sqrt{9.13}=\sqrt{117}\)
Vì \(\sqrt{117}>\sqrt{116}\) nên \(3\sqrt{13}>2\sqrt{29}\)
b)
Có:
\(\dfrac{5}{4}\sqrt{2}=\sqrt{\dfrac{25}{16}.2}=\sqrt{\dfrac{25}{8}}\)
\(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}=\sqrt{\dfrac{9}{4}.\dfrac{3}{2}}=\sqrt{\dfrac{27}{8}}\)
Do \(\sqrt{\dfrac{27}{8}}>\sqrt{\dfrac{25}{8}}\) nên \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}>\dfrac{5}{4}\sqrt{2}\)
c)
Có:
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\)
\(4\sqrt{3}=\sqrt{16.3}=\sqrt{48}\)
Vì \(\sqrt{50}>\sqrt{48}\) nên \(5\sqrt{2}>4\sqrt{3}\)
d)
Có:
\(\dfrac{5}{2}\sqrt{\dfrac{1}{6}}=\sqrt{\dfrac{25}{4}.\dfrac{1}{6}}=\sqrt{\dfrac{25}{24}}\)
\(6\sqrt{\dfrac{1}{37}}=\sqrt{36.\dfrac{1}{37}}=\sqrt{\dfrac{36}{37}}\)
lại có: \(\dfrac{25}{24}>\dfrac{36}{37}\)
\(\Rightarrow\dfrac{5}{2}\sqrt{\dfrac{1}{6}}>6\sqrt{\dfrac{1}{37}}\)
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: \(\left(\sqrt{2003}+\sqrt{2005}\right)^2=4008+2\cdot\sqrt{2003\cdot2005}=4008+2\cdot\sqrt{2004^2-1}\)
\(\left(2\sqrt{2004}\right)^2=4\cdot2004=4008+2\cdot\sqrt{2004^2}\)
=>\(\left(\sqrt{2003}+\sqrt{2005}\right)^2< \left(2\sqrt{2004}\right)^2\)
=>\(\sqrt{2003}+\sqrt{2005}< 2\sqrt{2004}\)
bài 45:so sánh
a)3\(\sqrt{3}\) và \(\sqrt{12}\)
b)7 và 3\(\sqrt{5}\)
c)\(\dfrac{1}{3}\sqrt{51}\) và \(\dfrac{1}{5}\sqrt{150}\)
d)\(\dfrac{1}{2}\sqrt{6}\) và \(6\sqrt{\dfrac{1}{2}}\)
a) \(3\sqrt{3}=\sqrt{27}>\sqrt{12}\)
b) \(3\sqrt{5}=\sqrt{45}>\sqrt{27}\)
c) \(\dfrac{1}{3}\sqrt{51}=\sqrt{\dfrac{51}{9}}< \sqrt{\dfrac{54}{9}}=6=\sqrt{\dfrac{150}{25}}=\dfrac{1}{5}\sqrt{150}\)
d) \(\dfrac{1}{2}\sqrt{6}=\sqrt{\dfrac{6}{4}}=\sqrt{\dfrac{3}{2}}< \sqrt{\dfrac{36}{2}}=6\sqrt{\dfrac{1}{2}}\)
Bài tập:so sánh
a. \(2\sqrt{3}\) và \(3\sqrt{2}\)
b. \(2\sqrt{3}+1\)và 4
c.\(\sqrt{2015}-\sqrt{2014}\) và \(\sqrt{2014}-\sqrt{2013}\)
a) Ta có: \(2\sqrt{3}=\sqrt{4\cdot3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{9\cdot2}=\sqrt{18}\)
mà \(\sqrt{12}< \sqrt{18}\)(vì 12<18)
nên \(2\sqrt{3}< 3\sqrt{2}\)
b) Ta có: \(\left(2\sqrt{3}+1\right)^2=8+4\sqrt{3}+1=9+4\sqrt{3}\)
\(4^2=16=9+7\)
mà \(4\sqrt{3}< 7\left(\sqrt{48}< \sqrt{49}\right)\)
nên \(\left(2\sqrt{3}+1\right)^2< 4^2\)
hay \(2\sqrt{3}+1< 4\)
c) Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)
\(\sqrt{2014}-\sqrt{2013}=\dfrac{1}{\sqrt{2014}+\sqrt{2013}}\)
Ta có: \(\sqrt{2015}+\sqrt{2014}>\sqrt{2013}+\sqrt{2014}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{2015}+\sqrt{2014}}< \dfrac{1}{\sqrt{2013}+\sqrt{2014}}\)
hay \(\sqrt{2015}-\sqrt{2014}< \sqrt{2014}-\sqrt{2013}\)
\(a\))Ta có:\(2\sqrt{3}=\sqrt{12}\)
\(3\sqrt{2}=\sqrt{18}\)
Vì \(\sqrt{12}< \sqrt{18}\)
⇒\(2\sqrt{3}< 3\sqrt{2}\)
\(b\))Ta có:\(2\sqrt{3}+1=\sqrt{12}+1\)
\(4=3+1=\sqrt{9}+1\)
Vì \(\sqrt{12}+1>\sqrt{9}+1\)
⇒\(2\sqrt{3}+1>4\)
So sánh:
a) \(1-\sqrt{3}\&\sqrt{0,2}\)
b) \(\sqrt{0,5}\&\sqrt{3}-2\)
c) \(\sqrt{2015}+\sqrt{2018}\&\sqrt{2016}+\sqrt{2017}\)
Ta có: \(\hept{\begin{cases}\sqrt{0,2}>0\\1=\sqrt{1}< \sqrt{3}\Rightarrow1-\sqrt{3}< 0\end{cases}\Rightarrow1-\sqrt{3}< \sqrt{0,2}}\)
Ta có: \(\hept{\begin{cases}\sqrt{0,5}>0\\\sqrt{3}< \sqrt{4}=2\Rightarrow\sqrt{3}-2< 0\end{cases}\Rightarrow\sqrt{0,5}>\sqrt{3}-2}\)
Không dùng mtct, so sánh
A) \(\sqrt{65}\)+1 và \(\sqrt{63}\)+1
B)\(\dfrac{1}{\sqrt{8}}\)và \(\dfrac{1}{\sqrt{7}}\)
C)\(\sqrt{34,9}\) và 6
D) \(3\sqrt{25,5}\) và 14
E)\(2\sqrt{26}\)+4 và 13
F) \(\sqrt{24}\)+\(\sqrt{63+3}\)và 16
G) \(\dfrac{46-3\sqrt{49}}{4}\)và \(\sqrt{50}\)
e: \(2\sqrt{26}>9\)
nên \(2\sqrt{26}+4>13\)
So sánh \(\sqrt{2015}+\sqrt{2018}\) và \(\sqrt{2016}+\sqrt{2017}\)
Ta có: \(\left(\sqrt{2015}+\sqrt{2018}\right)^2=4033+2\sqrt{2015.2018}\)
\(\left(\sqrt{2016}+\sqrt{2017}\right)^2=4033+2\sqrt{2016.2017}\)
\(2015.2018=2015.2017+2015=2017\left(2015+1\right)-2017+2015=2017.2016-2\)\(\Rightarrow2015.2018< 2016.2017\)
\(\Rightarrow4033+2\sqrt{2015.2018}< 4033+2\sqrt{2016.2017}\)
\(\Rightarrow\sqrt{2015}+\sqrt{2018}< \sqrt{2016}+\sqrt{2017}\left(đpcm\right)\)
Đặt \(A=\sqrt{2015}+\sqrt{2018}\Rightarrow A^{^2}=4033+2\sqrt{2015.2018}\)
\(B=\sqrt{2016}+\sqrt{2017}\Rightarrow B^{^2}=4033+2\sqrt{2016.2017}\)
Ta có: 2015.2018 = 2015.2017 + 2015
2016.2017 = 2015.2017 + 2017
Dễ dàng thấy được 2015.2018 < 2016.2017 => A2 < B2
=> A < B
Để phần so sánh chặt chẽ hơn, bạn có thể dùng cách này.
a, tính GT của đa thức \(f\left(x\right)=\left(x^4-3x+1\right)^{2016}\) tại \(x=9-\dfrac{1}{\sqrt{\dfrac{9}{4}-\sqrt{5}}}+\dfrac{1}{\sqrt{\dfrac{9}{4}+\sqrt{5}}}\)
b, so sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}và\dfrac{2.2016}{\sqrt{2017^2-1}-\sqrt{2016^2-1}}\)
c, tính GTBT: \(sinx.cosx+\dfrac{sin^2x}{1+cotx}+\dfrac{cos^2x}{1+tanx}\)
d, biết \(\sqrt{5}\) là số hữu tỉ, hãy tìm các số nguyên a,b tm::
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\)
a.
\(x=9-\dfrac{1}{\sqrt{\dfrac{9-4\sqrt{5}}{4}}}+\dfrac{1}{\sqrt{\dfrac{9+4\sqrt{5}}{4}}}\\ x=9-\dfrac{1}{\dfrac{\sqrt{5}-2}{2}}+\dfrac{1}{\dfrac{\sqrt{5}+2}{2}}\\ x=9-\left(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}\right)=9-8=1\\ \Rightarrow f\left(x\right)=f\left(1\right)=\left(1-1+1\right)^{2016}=1\)
c.
\(=\sin x\cdot\cos x+\dfrac{\sin^2x}{1+\dfrac{\cos x}{\sin x}}+\dfrac{\cos^2x}{1+\dfrac{\sin x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}+\dfrac{\cos^2x}{\dfrac{\sin x+\cos x}{\cos x}}\\ =\sin x\cdot\cos x+\dfrac{\sin^3x}{\sin x+\cos x}+\dfrac{\cos^3x}{\sin x+\cos x}\\ =\sin x\cdot\cos x+\dfrac{\left(\sin x+\cos x\right)\left(\sin^2x-\sin x\cdot\cos x+\cos^2x\right)}{\sin x+\cos x}\\ =\sin x\cdot\cos x-\sin x\cdot\cos x+\sin^2x+\cos^2x\\ =1\)
d.
\(\dfrac{2}{a+b\sqrt{5}}-\dfrac{3}{a-b\sqrt{5}}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{-a-5b\sqrt{5}}{\left(a+b\sqrt{5}\right)\left(a-b\sqrt{5}\right)}=-9-20\sqrt{5}\\ \Leftrightarrow\dfrac{a+5b\sqrt{5}}{a^2-5b^2}=9+20\sqrt{5}\\ \Leftrightarrow\left(9+20\sqrt{5}\right)\left(a^2-5b^2\right)=a+5b\sqrt{5}\\ \Leftrightarrow9\left(a^2-5b^2\right)+\sqrt{5}\left(20a^2-100b^2\right)-5b\sqrt{5}=a\\ \Leftrightarrow\sqrt{5}\left(20a^2-100b^2-5b\right)=9a^2-45b^2+a\)
Vì \(\sqrt{5}\) vô tỉ nên để \(\sqrt{5}\left(20a^2-100b^2-5b\right)\) nguyên thì
\(\left\{{}\begin{matrix}20a^2-100b^2-5b=0\\9a^2-45b^2+a=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}180a^2-900b^2-45b=0\\180a^2-900b^2+20a=0\end{matrix}\right.\\ \Leftrightarrow20a+45b=0\\ \Leftrightarrow4a+9b=0\Leftrightarrow a=-\dfrac{9}{4}b\\ \Leftrightarrow9a^2-45b^2+a=\dfrac{729}{16}b^2-45b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow\dfrac{9}{16}b^2-\dfrac{9}{4}b=0\\ \Leftrightarrow b\left(\dfrac{9}{16}b-\dfrac{9}{4}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}b=0\\b=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=9\end{matrix}\right.\)
Với \(\left(a;b\right)=\left(0;0\right)\left(loại\right)\)
Vậy \(\left(a;b\right)=\left(9;4\right)\)
Bài 1: rút gọn
\(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)\(\sqrt{\sqrt{3-\sqrt{3-\sqrt{13-4\sqrt{3}}}}}\)Bài 2: so sánh
x = \(\sqrt{2014}+\sqrt{2016}\)và y = \(2\sqrt{2015}\)m = \(\sqrt{2018}-\sqrt{2016}\)và n = \(\sqrt{2017}-\sqrt{2015}\)Bài 3: tìm x để căn thức có nghĩa
\(\sqrt{2x^2+5}\)\(\sqrt{2x^2-5}\)