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TD
12 tháng 2 2022
\(x^2+3x+1\)
=\(\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{5}{4}\)
=\(\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\)
Ta có:\(\left(x+\dfrac{3}{2}\right)^2\ge0\) Với mọi x
=>\(\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\)
Dấu "=" xảy ra <=>\(\left(x+\dfrac{3}{2}\right)^2=0\)
<=>\(x+\dfrac{3}{2}=0\)
<=>\(x=\dfrac{-3}{2}\)
LT
1
18 tháng 5 2022
\(x>0\)
\(C=x+\dfrac{1}{4x}+\dfrac{x}{\left(2x+1\right)^2}=\dfrac{4x^2+1}{4x}+\dfrac{x}{\left(2x+1\right)^2}\)
-Ta đặt \(A=T=4x^2+1;B=4x\) thì ta có:
\(A\ge B\Rightarrow A+T\ge B+T\) (do \(T>0\))\(\Rightarrow\dfrac{A+T}{B+T}\ge1\)
-Do đó: \(C=\dfrac{4x^2+1}{4x}+\dfrac{x}{\left(2x+1\right)^2}\ge\text{}\dfrac{4x^2+1+4x^2+1}{4x+4x^2+1}+\dfrac{x}{\left(2x+1\right)^2}=\dfrac{2\left(4x^2+1\right)}{\left(2x+1\right)^2}+\dfrac{8x}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=\dfrac{2\left(2x+1\right)^2}{\left(2x+1\right)^2}-\dfrac{7x}{\left(2x+1\right)^2}=2-\dfrac{7x}{\left(2x+1\right)^2}\)
-Áp dụng BĐT AM-GM ta có:
\(C\ge2-\dfrac{7x}{\left(2x+1\right)^2}\ge2-\dfrac{7x}{4.2x}=2-\dfrac{7}{8}=\dfrac{9}{8}\)
\(C=\dfrac{9}{8}\Leftrightarrow x=\dfrac{1}{2}\)
-Vậy \(C_{min}=\dfrac{9}{8}\)
NM
0
HT
0
SM
1
F
26 tháng 9 2020
\(y-x=1\Rightarrow x=y-1\)
\(\Rightarrow x^2+y^2=\left(y-1\right)^2+y^2\)
\(=y^2-2y+1+y^2\)
\(=2y^2-2y+1\)
\(=2\left(y^2-y+\frac{1}{2}\right)\)
\(=2\left(y^2-2y\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{2}\)
\(=2\left(y-\frac{1}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}\forall y\)
Dấu"=" xảy ra khi \(2\left(y-\frac{1}{2}\right)^2=0\Rightarrow y=\frac{1}{2}\)
Vì \(y-x=1\)nên
\(\Rightarrow\frac{1}{2}-x=1\Rightarrow x=-\frac{1}{2}\)
Vậy \(Min_A=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2};y=\frac{1}{2}\)
UK
3 tháng 3 2019
\(\dfrac{x^2}{1+x^4}\ge\dfrac{0}{1+x^4}=0\)
GTNN là 0 khi x=0
\(\dfrac{x^2}{1+x^4}\le\dfrac{1}{2}\Leftrightarrow\left(x^2-1\right)^2\ge0\)
GTLN là \(\dfrac{1}{2}\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
TN
1
T
17 tháng 7 2019
\(A=\left(x+3\right)\left(x-4\right)+7=x^2-x-5=\left(x^2-x+\frac{1}{4}\right)-\frac{1}{4}-5\)
\(=\left(x-\frac{1}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)
"=" <=> x = 1/2
\(B=3-\left(x-1\right)\left(x-2\right)=3-\left(x^2-3x+2\right)\)
\(=3-\left(x-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+2\right)\)
\(=3+\frac{1}{4}-\left(x-\frac{3}{2}\right)^2\le\frac{13}{4}\)
Xảy ra khi x = 3/2
Có : \(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)\(\ge\left|x^2-x+1-x^2+x-2\right|=\left|-1\right|=1\)
Vậy Pmin=1\(\Leftrightarrow\left(x^2-x+1\right)\left(-x^2+x-2\right)\ge0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2-x+2\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-x+1\ge0\\x^2-x+2\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-x+1\le0\\x^2-x+2\ge0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\in R\\x\in\varnothing\end{matrix}\right.\\\left\{{}\begin{matrix}x\in\varnothing\\x\in R\end{matrix}\right.\end{matrix}\right.\)
Vậy không tồn tại GTNN của P.
\(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)
\(P=\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right|+\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{7}{4}\right|\)
\(P=\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right|+\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right|\)
\(P=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{10}{4}\ge\dfrac{10}{4}=\dfrac{5}{2}\)
\(\Rightarrow P_{min}=\dfrac{5}{2}\) khi \(x=\dfrac{1}{2}\)