gtnn cua p=(x-1)(2x+3)
tìm gtnn cua:|2x-3|+2|x-1|
Đặt \(A=\left|2x-3\right|+2\left|x-1\right|\)
\(\Rightarrow A=\left|2x-3\right|+\left|2x-2\right|=\left|2x-3\right|+\left|2-2x\right|\)
\(\Rightarrow A\ge\left|2x-3+2-2x\right|=\left|-1\right|=1\)
Dấu " = " xảy ra \(\Leftrightarrow\left(2x-3\right)\left(2-2x\right)\ge0\)\(\Leftrightarrow\left(2x-3\right)\left(1-x\right)\ge0\)
TH1: \(\hept{\begin{cases}2x-3\le0\\1-x\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\1\le x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le\frac{3}{2}\\x\ge1\end{cases}}\Leftrightarrow1\le x\le\frac{3}{2}\)
TH2: \(\hept{\begin{cases}2x-3\ge0\\1-x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\1\ge x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge\frac{3}{2}\\x\le1\end{cases}}\)( vô lý )
Vậy \(minA=1\Leftrightarrow1\le x\le\frac{3}{2}\)
tim gtnn cua da thuc P=(x-1)(2x+3)
Ta có :
\(P=\left(x-1\right)\left(2x+3\right)=2x^2-2x+3x-3\) \(=2x^2+x-3\)
\(=2\left(x^2+\frac{1}{2}x-\frac{3}{2}\right)\) \(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}\right)\)
\(=2\left(x^2+\frac{1}{2}x+\frac{1}{16}-\frac{23}{16}\right)\)
\(=2\left(x+\frac{1}{4}\right)^2-\frac{23}{8}\ge-\frac{23}{8},\)với mọi x
Vậy \(MIN_P=\frac{-23}{8}\) khi \(x+\frac{1}{4}=0\Leftrightarrow x=\frac{-1}{4}\)
voi x > 1/2
tim gtnn cua D=x/3 + 5/2x-1
GTNN cua x^2-2x-3
x2-2x-3=(x-1)2-4>-4
Dấu "=" xảy ra <=> x=1
B=/2x-100/ +/200-2x/ tìm GTNN cua B
C=/x-70/+/20+x/ timf GTNN cua C
D=/x-80/+/x-200/ timfGTTN cua D
ap dung bdt co si tim gtnn cua bieu thuc y=x/3 +5/2x+1;x>1/2
Tim gtnn cua 2x/(x-1)^2 khi x>1
x=3 thì Min là 9 nha bạn
vậy nếu x=5 thì sao nhỉ
tim GTNN cua C=2x+1/x^2
Tìm gtln cua bt:
A= (2x+1/2)⁴-1
Tìm gtnn cua bt
B=-(4/9 .x - 2/15)²+3