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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 <=>\(\left(x+\dfrac{3}{2}\right)^2=0\)

                        <=>\(x+\dfrac{3}{2}=0\)

                        <=>\(x=\dfrac{-3}{2}\)

 

12 tháng 2 2022

min =1 

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

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 \(C_{min}=\dfrac{9}{8}\)

1 tháng 3 2019

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.

NV
3 tháng 3 2019

\(P=\left|x^2-x+1\right|+\left|x^2-x+2\right|\)

\(P=\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\right|+\left|x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{7}{4}\right|\)

\(P=\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right|+\left|\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right|\)

\(P=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{10}{4}\ge\dfrac{10}{4}=\dfrac{5}{2}\)

\(\Rightarrow P_{min}=\dfrac{5}{2}\) khi \(x=\dfrac{1}{2}\)

21 tháng 10 2021

\(A=\left(x-4\right)\left(x+3\right)\)

\(=x^2-x-12\)

\(=x^2-x+\dfrac{1}{4}-\dfrac{49}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2-\dfrac{49}{4}\ge-\dfrac{49}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

 

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 \(\dfrac{1}{2}\Leftrightarrow\)\(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

13 tháng 9 2015

x^2 - 4x + 1 = x^2 - 4x + 4 - 3 = ( x- 2 )^2 - 3 

Vậy GTnn là 3 khi x = 2 

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 \(Min_A=\frac{1}{2}\Leftrightarrow x=-\frac{1}{2};y=\frac{1}{2}\)