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YH
2
K
1
HP
20 tháng 2 2021
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):
\(A=\left|x-3\right|+\left|x-1\right|+\left|x+1\right|+\left|x+3\right|\)
\(=\left|3-x\right|+\left|x+3\right|+\left|1-x\right|+\left|x+1\right|\)
\(\ge\left|3-x+x+3\right|+\left|1-x+x+1\right|=8\)
\(minA=8\Leftrightarrow\left\{{}\begin{matrix}\left(3-x\right)\left(x+3\right)\ge0\\\left(1-x\right)\left(x+1\right)\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le1\)
23 tháng 1 2021
- Ta có đồ thị hàm số :
- Theo đồ thị hàm số : Min = 0 tại x = 0 .
UK
4 tháng 12 2017
a) \(\dfrac{\left(x-1\right)^2}{x-2}=\dfrac{\left(x-2\right)^2+2\left(x-2\right)+1}{x-2}=x-2+2+\dfrac{1}{x-2}\ge2+2\sqrt{\left(x-2\right).\dfrac{1}{x-2}}=4\)
GTNN là 4 khi x=3
3 tháng 12 2023
a: \(f\left(x\right)=2x^2-7x+9\)
=>\(f'\left(x\right)=2\cdot2x-7=4x-7\)
Đặt f'(x)=0
=>\(4x-7=0\)
=>\(x=\dfrac{7}{4}\)
\(f\left(\dfrac{7}{4}\right)=2\cdot\left(\dfrac{7}{4}\right)^2-7\cdot\dfrac{7}{4}+9=\dfrac{23}{8}\)
\(f\left(-1\right)=2\left(-1\right)^2-7\cdot\left(-1\right)+9=18\)
\(f\left(4\right)=2\cdot4^2-7\cdot4+9=13\)
Vì \(f\left(\dfrac{7}{4}\right)< f\left(4\right)< f\left(-1\right)\)
nên \(f\left(x\right)_{max\left[-1;4\right]}=18;f\left(x\right)_{min\left[-1;4\right]}=\dfrac{23}{8}\)
b: \(f\left(x\right)=x^2+5x+3\)
=>\(f'\left(x\right)=2x+5\)
f'(x)=0
=>2x+5=0
=>2x=-5
=>\(x=-\dfrac{5}{2}\)
\(f\left(-\dfrac{5}{2}\right)=\left(-\dfrac{5}{2}\right)^2+5\cdot\dfrac{-5}{2}+3=\dfrac{25}{4}-\dfrac{25}{2}+3=-\dfrac{13}{4}\)
\(f\left(2\right)=2^2+5\cdot2+3=4+10+3=17\)
\(f\left(6\right)=6^2+5\cdot6+3=69\)
Vậy: \(f\left(x\right)_{max\left[2;6\right]}=69;f\left(x\right)_{min\left[2;6\right]}=-\dfrac{13}{4}\)
KK
21 tháng 1 2017
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
HN
25 tháng 11 2018
\(P=3\left(x^2+y^2\right)^2-3x^2y^2-2\left(x^2+y^2\right)+1\)
\(\ge3\left(x^2+y^2\right)^2-\dfrac{3}{4}\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1\)
Đặt \(x^2+y^2=a\) thì \(a\ge2\).Xét hàm \(f\left(a\right)=\dfrac{9}{4}a^2-2a+1\)
Dế thấy \(f_{(a)}\) đồng biến trên [2,+\(\infty\)] nên \(f_{Min}\)=\(f_{(2)}\)=6
Dấu = xảy ra khi x=y=1
