GIải giúp mk câu này với
Biết a+b=1. Hãy chứng minh 3a2 + b2 >= 3/4
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3\pi}{2}\le a\le2\pi\Rightarrow3\pi\le2a\le4\pi\)
\(\Rightarrow sin2a\le0\)
\(cos^2a-sin^2a=\dfrac{1}{2}\Leftrightarrow cos2a=\dfrac{1}{2}\)
\(\Rightarrow sin2a=-\sqrt{1-cos^22a}=-\dfrac{\sqrt{3}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
Ta có: M = a2+ab+b2 -3a-3b-3a-3b +2001
=> 2M = ( a2 + 2ab + b2) -4.(a+b) +4 + (a2 -2a+1)+(b2 -2b+1) + 3996
2M= ( a+b-2)2 + (a-1)2 +(b-1)2 + 3996
=> MinM = 1998 tại a=b=1
Câu 3:
Ta có: P= x2 +xy+y2 -3.(x+y) + 3
=> 2P = ( x2 + 2xy +y2) -4.(x+y) + 4 + (x2 -2x+1) +(y2 -2y+1)
2P = ( x+y-2)2 +(x-1)2+(y-1)2
=> MinP = 0 tại x=y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{3a^2-b^2}{a^2+b^2}=\dfrac{3}{4}\)
\(\Leftrightarrow4\cdot\left(3a^2-b^2\right)=3\left(a^2+b^2\right)\)
\(\Leftrightarrow12a^2-4b^2=3a^2+3b^2\)
\(\Leftrightarrow12a^2-3a^2=3b^2+4b^2\)
\(\Leftrightarrow9a^2=7b^2\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{7}{9}\)
hay \(\dfrac{a}{b}=\pm\dfrac{\sqrt{7}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bai 1
=ax5-x5-9xy-4xy-7x
=ax5-(5x+7x)-(9xy+4xy)
=5ax-12x-13xy
2
M=4a+ab-2b+2a-2b+ab
=6a+2ab-4b
n=6a+2b-ab+2a
=8a+2b-ab
m-n=6a+2ab-4b-8a-2b+ab
=3ab-2a-6b
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{a}{b}=\dfrac{b}{c}\Rightarrow ac=b^2\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)
\(a+b=1\Rightarrow b=1-a\Rightarrow b^2=\left(1-a\right)^2\)
\(\Rightarrow3a^2+b^2=3a^2+\left(1-a\right)^2=4a^2-2a+1\)
Mà \(4a^2-2a+1=\left(2a\right)^2-2.2a.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(2a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)\(\left(đpcm\right)\)