A | B |
1. I'm cold |
a. He'd like to sit down |
2. Nam is tired |
b. She'd like some water |
3. He's hungry |
c. I want a hot drink |
4. Hoa's thirsty |
d. I want some orange juice |
5. I'm full |
e. He'd like some noodles |
6. I'm hot |
f. I don't want any noodles |
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![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(VT=\dfrac{1}{a+1}+\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{1}{a}\)=VP
b: \(VP=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}=VP\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(=\dfrac{-x-1+2x-2-x+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
b: Để A>0 thì 1-2x>0
=>2x<1
=>x<1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{6a+1}{3a-1}=2+\frac{3}{3a-1}\)
Để (6a+1) ⋮ (3a -1) thì: 3a-1 thuộc Ư(3) ={1; -1; 3; -3}
-Với 3a-1=1 => a=\(\frac{2}{3}\) (Loại)
- Với 3a- 1= -1 => a= 0 (Chọn)
- Với 3a -1 = 3 => a= \(\frac{4}{3}\)(Loại)
- Với 3a- 1= -3=> a= \(\frac{-2}{3}\)( Loại)
Vậy số nguyên a cần tìm là 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Có \(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
<=> \(a+b+c\ge3.\frac{bc+ac+ab}{abc}=\frac{bc+ac+ab}{a+b+c}\)( vì abc=a+b+c)
<=> \(\left(a+b+c\right)^2\ge3\left(bc+ac+ab\right)\)
<=> \(a^2+b^2+c^2+2bc+2ac+2ab-3bc-3ac-3ab\ge0\)
<=> \(a^2+b^2+c^2-ab-ac-bc\ge0\)
<=> 2a2+2b2+2c2-2ab-2ac-2bc \(\ge0\)
<=> (a2-2ab+b2)+(b2-2bc+c2)+(c2-2ac+a2) \(\ge0\)
<=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.< =>a=b=c\)
Vậy \(a+b+c\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
1) Có \(2x=-3y=4z\)
=> \(y=\frac{2x}{-3}\) ,\(z=\frac{2x}{4}=\frac{x}{2}\)
Có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
<=> \(\frac{1}{x}+\frac{1}{\frac{2x}{-3}}+\frac{1}{\frac{x}{2}}=3\)
<=>\(\frac{1}{x}-\frac{3}{2x}+\frac{2}{x}=3\) <=> \(\frac{2-3+4}{2x}=3\) <=> 3=6x
<=> x=\(\frac{1}{2}\)
=> y=\(\frac{\frac{1}{2}.2}{-3}=-\frac{1}{3}\) , \(z=\frac{2}{\frac{1}{2}}=4\)
Vậy (x,y,z)\(\in\left\{\frac{1}{2},-\frac{1}{3},4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/CM: \(\left(\frac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}\)
\(\Leftrightarrow a+b\ge2\sqrt{ab}\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng với mọi a,b>0)
CM: \(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)
\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{4}\ge\frac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2\ge2ab\) ( luôn đúng)
b/CM: \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)
\(\Leftrightarrow\frac{4\left(a^3+b^3\right)}{8}\ge\frac{\left(a+b\right)^3}{8}\)
\(\Leftrightarrow3\left(a^3+b^3\right)\ge3a^2b+3ab^2\)
\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) ( luôn đúng với mọi a,b>0)
c/CM: \(a^4+b^4\ge a^3b+ab^3\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2+ab\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+\frac{2ab}{2}+\frac{b^2}{4}+\frac{3b^2}{4}\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right)\ge0\) ( luôn đúng)
d/Ta xét hiệu: \(a^4-4a+3\)
\(=a^4-2a^2+1+2a^2-4a+2\)
\(=\left(a-1\right)^2+2\left(a-1\right)^2\ge0\)
Suy ra BĐT luôn đúng
e/Ta xét hiệu:( Làm nhanh)
\(a^3+b^3+c^3-3abc\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=\frac{1}{2}\left(a+b+c\right)\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\ge0\)
f/Ta có: \(\frac{a^6}{b^2}-a^4+\frac{a^2b^2}{4}+\frac{b^6}{a^2}-b^4+\frac{a^2b^2}{4}\)
\(=\left(\frac{a^3}{b}-\frac{ab}{2}\right)^2+\left(\frac{b^3}{a}-\frac{ab}{2}\right)^2\ge0\)(1)
Mà \(\frac{a^2b^2}{4}+\frac{a^2b^2}{4}\ge0\)(2)
Lấy (1) trừ (2) được: \(\frac{a^6}{b^2}+\frac{b^6}{a^2}-a^4-b^4\ge0\RightarrowĐPCM\)
g/Làm rồi..xem lại trong trang cá nhân
h/Xét hiệu có: \(\left(a^5+b^5\right)\left(a+b\right)-\left(a^4+b^4\right)\left(a^2+b^2\right)\)
\(=a^5b+ab^5-a^2b^4-a^4b^2\)
\(=a^4b\left(a-b\right)-ab^4\left(a-b\right)\)
\(=ab\left(a^2-b^2\right)\left(a-b\right)\)
\(=ab\left(a+b\right)\left(a-b\right)^2\ge0\forall ab>0\)
Suy ra ĐPCM
1c 2a 3e 4b 5f 6d
a. He'd like to sit down
b. She'd like some water
c. I want a hot drink
d. I want some orange juice
e. He'd like some noodles
f. I don't want any noodles
1. c
2. a
3. e
4. b
5. f
6. d