\(A=\left(4x^2-4x+1\right)+\left(x+\frac{1}{4x}\right)+2013\ge2013+1=2014;;;.\)

A min = 2014 khi x =1/2 

b) B= 5x² -10x+3-2

B = (5x² - 2.5.1 . 1²)-2

B = (5x-1)²-2 

ta có :

(5x-1)² > 0 với mọi x thuộc R

(5x-1)² -2 < -2

vậy B < -2

dấu = xảy ra <=> x = 1/5

mai tui lm nốt choa

a)

\(A=4x^2-4x-1=4x^2-4x+1-2=\left(2x-1\right)^2-2\)

\(A\ge-2\forall x\in R\) 

Dấu "=" xảy ra <=>\(\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\) 

Vậy A_min =-2 tại x=1/2

M=(4x²-4x+1)+(x+\(\dfrac{1}{4x}\))+2013

=(2x-1)²+(x+\(\dfrac{1}{4x}\))+2013

x>0 nên áp dụng BĐT côsi cho 2 số không âm:

\(x+\dfrac{1}{4x}\ge2\sqrt{\dfrac{x}{4x}}=1\)

Dấu "=" xảy ra khi 4x²=1<=>x=\(\dfrac{1}{2}\)

(2x-1)²\(\ge\)0 với mọi x

Dấu "=" xảy ra khi x=\(\dfrac{1}{2}\)

=>M\(\ge\)0+1+2013=2014

=>M_min=2014 khi và chỉ khi x=\(\dfrac{1}{2}\)

Vậy...

Bài 1:

a) \(x^2-5x+1=0\)

\(\Leftrightarrow\left(x^2-5x+\frac{25}{4}\right)-\frac{21}{4}=0\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{\left(\sqrt{21}\right)^2}{2^2}=0\)

\(\Leftrightarrow\left(x-\frac{5+\sqrt{21}}{2}\right)\left(x+\frac{\sqrt{21}-5}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5+\sqrt{21}}{2}=0\\x+\frac{\sqrt{21}-5}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{cases}}\)

b) \(3x^2-12x-1=0\)

\(\Leftrightarrow3\left(x^2-4x+4\right)-13=0\)

\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{\frac{13}{3}}\right)^2=0\)

\(\Leftrightarrow\left(x-2-\sqrt{\frac{13}{3}}\right)\left(x-2+\sqrt{\frac{13}{3}}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2+\sqrt{\frac{13}{3}}\\x=2-\sqrt{\frac{13}{3}}\end{cases}}\)

Bài 2:

a) \(A=\frac{1}{4}x^2-x+1=\left(\frac{1}{2}x-1\right)^2\ge0\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\left(\frac{1}{2}x-1\right)^2=0\Rightarrow\frac{1}{2}x=1\Rightarrow x=2\)

Vậy Min(A) = 0 khi x = 2

b) \(B=3x^2-4x-2=3\left(x^2-\frac{4}{3}x+\frac{4}{9}\right)-\frac{10}{3}=3\left(x-\frac{2}{3}\right)^2-\frac{10}{3}\ge-\frac{10}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(3\left(x-\frac{2}{3}\right)^2=0\Rightarrow x=\frac{2}{3}\)

Vậy \(Min\left(B\right)=-\frac{10}{3}\Leftrightarrow x=\frac{2}{3}\)

Giúp mk vs các bn eii

\(P=-\left(4x^2-4x+1+x+\frac{1}{4x}-2015\right)\)

\(=-\left[\left(2x-1\right)^2+\frac{\left(2x-1\right)^2}{4x}\right]+2014\)

\(P\le2014\forall x>0\)

Dấu "=" xảy ra <=> x=\(\frac{1}{2}\)

Lời giải:

a)

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

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

Với \(x\geq \frac{3}{2}\Rightarrow \left\{\begin{matrix} 2x+1>0\\ 2x-3\geq 0\end{matrix}\right.\Rightarrow A=(2x+1)(2x-3)+4\geq 4\)

Vậy GTNN của $A$ là $4$ khi $x=\frac{3}{2}$

b)

\(B=5x^2-10x+3=5(x^2-2x+1)-2\)

\(=5(x-1)^2-2\)

Ta thấy \((x-1)^2\geq 0, \forall x\geq 1\Rightarrow B=5(x-1)^2-2\geq -2\)

Vậy GTNN của $B$ là $-2$ khi $(x-1)^2=0\Leftrightarrow x=1$

c)

\(C=4x^2-6x+2=(2x)^2-2.2x.\frac{3}{2}+(\frac{3}{2})^2-\frac{1}{4}\)

\(=(2x-\frac{3}{2})^2-\frac{1}{4}\)

Ta thấy \((2x-\frac{3}{2})^2\geq 0, \forall x\geq 0\Rightarrow C=(2x-\frac{3}{2})^2-\frac{1}{4}\geq -\frac{1}{4}\)

Vậy GTNN của $C$ là $\frac{-1}{4}$ khi \((2x-\frac{3}{2})^2=0\Leftrightarrow x=\frac{3}{4}\)

d)

\(D=3x^2+2x+1=3(x^2+\frac{2}{3}x+\frac{1}{9})+\frac{2}{3}\)

\(=3(x+\frac{1}{3})^2+\frac{2}{3}\)

Ta thấy \((x+\frac{1}{3})^2\geq 0, \forall x\geq -1\Rightarrow D=3(x+\frac{1}{3})^2+\frac{2}{3}\geq \frac{2}{3}\)

Vậy GTNN của $D$ là $\frac{2}{3}$ khi $(x+\frac{1}{3})^2=0\Leftrightarrow x=-\frac{1}{3}$

a: 

Sửa đề: \(P=\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{4x^2}{x^2-9}\right):\left(\dfrac{5}{3-x}-\dfrac{4x+2}{3x-x^2}\right)\)\(P=\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{4x^2}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{5x-4x-2}{x\left(3-x\right)}\)

\(=\dfrac{-x^2-6x-9+x^2-6x+9-4x^2}{\left(x-3\right)\left(x+3\right)}:\dfrac{x-2}{x\left(3-x\right)}\)

\(=\dfrac{-4x^2-12x}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(3-x\right)}{x-2}\)

\(=\dfrac{-4x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-x\left(x-3\right)}{x-2}=\dfrac{4x^2}{x-2}\)

b: x^2-4x+3=0

=>x=1(nhận) hoặc x=3(loại)

Khi x=1 thì \(P=\dfrac{4\cdot1^2}{1-2}=-4\)

c: P>0

=>x-2>0

=>x>2

d: P nguyên

=>4x^2 chia hết cho x-2

=>4x^2-16+16 chia hết cho x-2

=>x-2 thuộc {1;-1;2;-2;4;-4;8;-8;16;-16}

=>x thuộc {1;4;6;-2;10;-6;18;-14}