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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có B=\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<1
=>\(\frac{2009^{2010}-2}{2009^{2011}-2}\)<\(\frac{2009^{2010}-2+3}{2009^{2011}-2+3}\)=\(\frac{2009^{2010}+1}{2009^{2011}+1}\)(1)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<1
=> \(\frac{2009^{2010}+1}{2009^{2011}+1}\)<\(\frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}\)=\(\frac{2009^{2010}+2009}{2009^{2011}+2009}\)=\(\frac{2009\cdot\left(2009^{2009}+1\right)}{2009\cdot\left(2009^{2010}+1\right)}\)=\(\frac{2009^{2009}+1}{2009^{2010}+1}\)=A(2)
Từ (1)và(2)=>B<\(\frac{2009^{2010}+1}{2009^{2011}+1}\)<A=>B<A hay A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $A=\left\{1; 2; 3; 4; 5\right\}$
$B=\left\{3; 4; 5;6 ;7\right\}$
$A\cap B=\left\{ 3; 4;5\right\}$
$A\cup B =\left\{1;2 ;3; 4; 5;6 ;7\right\}$
b.
$A\setminus B = (-2;-1)$
![](https://rs.olm.vn/images/avt/0.png?1311)
`(x^2+1)/((3-x)(x+2))>=0(x ne -2,3)`
Vì `x^2+1>0`
`=>(3-x)(x+2)>0`
`=>(x-3)(x+2)<0`
`=>-2<x<3`
Ủa thì chọn gì?
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=1+2+2^2+2^3+...+2^{2010}\)
\(2A=2+2^2+2^3+2^4+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2011}\right)-\left(1+2+2^2+...+2^{2010}\right)\)
\(\Rightarrow A=2^{2011}-1\)
\(B=2^{2011}-1\)
Vậy A = B.
![](https://rs.olm.vn/images/avt/0.png?1311)
a3+b3+c3=3abc
<=>(a+b)3-3ab(a+b)-3abc+c3=0
<=>(a+b+c)[(a+b)2-(a+b)c+c2]-3ab.(a+b+c)=0
<=>(a+b+c)(a2+b2+c2-ab-bc-ac)=0
<=>(a+b+c)(2a2+2b2+2c2-2ab-2bc-2ac)=0
<=>(a+b+c)[(a-b)2+(b-c)2+(c-a)2]=0
<=>a+b+c=0 [(a-b)2+(b-c)2+(c-a)2 khác 0]
=>a2+b2-c2=-2ab;b2+c2-a2=-2bc;c2+a2-b2=-2ac
Suy ra : P=\(-\left(\dfrac{1}{2ab}+\dfrac{1}{2bc}+\dfrac{1}{2ac}\right)=-\dfrac{a+b+c}{2abc}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt \(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)
Ta có \(A=(a-\frac{ab^2}{1+b^2})+(b-\frac{bc^2}{1+c^2})+(c-\frac{ca^2}{1+a^2})=3-\left ( \frac{ab^2}{1+b^2}+\frac{bc^2}{1+c^2}+\frac{ca^2}{1+a^2} \right )\)
Áp dụng bất đẳng thức AM-GM:
\(A\geq 3-\left ( \frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{3a} \right )=3-\frac{1}{2}(ab+bc+ac)\)
Cũng theo AM-GM
\(9=(a+b+c)^2\geq 3(ab+bc+ac)\Rightarrow ab+bc+ac\leq 3\)
\(\Rightarrow A\geq 3-\frac{3}{2}=\frac{3}{2}\)
Dấu $=$ xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Bunhiacopxki:
\(\left(1^2+4^2\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(1.a+4.\frac{1}{b}\right)^2\)\(\Rightarrow a^2+\frac{1}{b^2}\ge\frac{1}{17}\left(a+\frac{4}{b}\right)^2\)
\(\Rightarrow\sqrt{a^2+\frac{1}{b^2}}\ge\frac{1}{\sqrt{17}}\left(a+\frac{4}{b}\right)\)
Tương tự, ta có: \(\sqrt{b^2+\frac{1}{c^2}}\ge\frac{1}{\sqrt{17}}\left(b+\frac{4}{c}\right)\)
và \(\sqrt{c^2+\frac{1}{a^2}}\ge\frac{1}{\sqrt{17}}\left(c+\frac{4}{a}\right)\)
Cộng từng vế của các BĐT trên, ta được:
\(P\ge\frac{1}{\sqrt{17}}\left(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)\)\(\ge\frac{1}{\sqrt{17}}\left(a+b+c+\frac{36}{a+b+c}\right)\)(svac - xơ)
\(=\frac{1}{\sqrt{17}}\left[\left(a+b+c\right)+\frac{9}{4\left(a+b+c\right)}+\frac{135}{4\left(a+b+c\right)}\right]\ge\frac{3\sqrt{17}}{2}\)
Vậy \(P=\sqrt{a^2+\frac{1}{b^2}}\)\(+\sqrt{b^2+\frac{1}{c^2}}\)\(+\sqrt{c^2+\frac{1}{a^2}}\ge\frac{3\sqrt{17}}{2}\)
(Dấu "="\(\Leftrightarrow a=b=c=2\))
Bài em làm ok rồi nhưng mà dấu bằng xảy ra bị sai. Em kiểm tra lại!๖²⁴ʱČøøℓ ɮøү 2к⁷༉
\(P=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}+\dfrac{1}{2^{20}}\\ 2P=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{18}}+\dfrac{1}{2^{19}}\\ 2P-P=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{18}}+\dfrac{1}{2^{19}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}+\dfrac{1}{2^{20}}\right)\\ P=1-\dfrac{1}{2^{20}}\)
gúp mình bài 2 thôi nha