biet a+b=4 Tinh GTLN cua A = \(2\left(a^3b+ab^3\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
gt <=> \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)
Đặt: \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
=> Thay vào thì \(VT=\frac{\frac{1}{xy}}{\frac{1}{z}\left(1+\frac{1}{xy}\right)}+\frac{1}{\frac{yz}{\frac{1}{x}\left(1+\frac{1}{yz}\right)}}+\frac{1}{\frac{zx}{\frac{1}{y}\left(1+\frac{1}{zx}\right)}}\)
\(VT=\frac{z}{xy+1}+\frac{x}{yz+1}+\frac{y}{zx+1}=\frac{x^2}{xyz+x}+\frac{y^2}{xyz+y}+\frac{z^2}{xyz+z}\ge\frac{\left(x+y+z\right)^2}{x+y+z+3xyz}\)
Có BĐT x, y, z > 0 thì \(\left(x+y+z\right)\left(xy+yz+zx\right)\ge9xyz\)Ta thay \(xy+yz+zx=1\)vào
=> \(x+y+z\ge9xyz=>\frac{x+y+z}{3}\ge3xyz\)
=> Từ đây thì \(VT\ge\frac{\left(x+y+z\right)^2}{x+y+z+\frac{x+y+z}{3}}=\frac{3}{4}\left(x+y+z\right)\ge\frac{3}{4}.\sqrt{3\left(xy+yz+zx\right)}=\frac{3}{4}.\sqrt{3}=\frac{3\sqrt{3}}{4}\)
=> Ta có ĐPCM . "=" xảy ra <=> x=y=z <=> \(a=b=c=\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+2x^2y^2+y^4-3x^2-4y^2+4=1\)
\(\Leftrightarrow\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+4=1-x^2\)
\(\Leftrightarrow\left(x^2+y^2-2\right)^2=1-x^2\)
Do \(1-x^2\le1\) \(\forall x\)
\(\Rightarrow-1\le x^2+y^2-2\le1\)
\(\Rightarrow1\le x^2+y^2\le3\)
\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)
\(A_{max}=3\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b=4
<=> a^2 + b^2 +2ab=16
<=> a^2 +b^2 = 16 - 2ab (1)
Ta có A=2(a^3.b + a.b^3)
<=> A= 2.a.b.(a^2 +b^2)
Từ (1) => A= 2a.b.(16 -2a.b)
<=> A=-4.a^2.b^2 + 32.a.b
<=> A = -4a^2.b^2 + 32a.b - 64 +64
<=> A= -4.(a.b - 4)^2 +64
.........
(Tự làm)