help me, please!!!!!!!!!!!!!!!

\(B=\frac{x^2-2x+2018}{2018x^2}\)

\(=\frac{1}{2018}-\frac{2}{2018x}+\frac{1}{x^2}\)

\(=\left(\frac{1}{x}-\frac{1}{\sqrt{2018}}\right)^2\ge0\)

Vậy giá trị nhỏ nhất \(B=0\)khi và chỉ khi  \(\frac{1}{x}-\frac{1}{\sqrt{2018}}=0\)

\(\Rightarrow\frac{1}{x}=\frac{1}{\sqrt{2018}}\)

\(\Rightarrow x=\sqrt{2018}\)

\(B=\frac{x^2-2x+2018}{2018x^2}\)

\(=\frac{2018x^2-2\cdot2018\cdot x+2018^2}{2018^2x^2}\)

\(=\frac{2017x^2+\left(x^2-2\cdot2018\cdot x+2018^2\right)}{2018^2x^2}\)

\(=\frac{2017}{2018^2}+\frac{\left(x-2018\right)^2}{2018^2x^2}\ge\frac{2017}{2018^2}\)

Đẳng thức xảy ra tại x=2018

Vậy .........

Bài 2 :

\(A=4x^2-2.2x.2+4+1\)

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

Thấy : \(\left(2x-2\right)^2\ge0\)

\(A=\left(2x-2\right)^2+1\ge1\)

Vậy \(MinA=1\Leftrightarrow x=1\)

\(B=\left(5x\right)^2-2.5x.1+1-4\)

\(=\left(5x-1\right)^2-4\)

Thấy : \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)

Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)

\(C=\left(7x\right)^2-2.7x.2+4-5\)

\(=\left(7x-2\right)^2-5\)

Thấy : \(\left(7x-2\right)^2\ge0\)

\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)

Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)

\(1.\)

\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)

\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)

\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5

\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)

\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)

\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)

dấu"=" xảy ra<=>x=1/4

a: A=x^2-6x+9+2=(x-3)^2+2>=2

Dấu = xảy ra khi x=3

b: B=x^2-20x+100+1=(x-10)^2+1>=1

Dấu = xảy ra khi x=10

d: C=x^2-16x+8+3

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

Dấu = xảy ra khi x=4

Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0

=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0 

Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1

Thay vào bt S :

S = ( 2 - 1)^2019 + (2-1)^2019

= 1^2019 + 1^2019 = 2

em cảm ơn

a,\(A=\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=\left(x^2+6x+5\right)\left(x^2+6x+8\right)\)

đặt \(x^2+6x+5=t=>t\left(t+3\right)=t^2+3t=t^2+2.\dfrac{3}{2}t+\dfrac{9}{4}-\dfrac{9}{4}\)

\(=\left(t+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}< =>t=\dfrac{-3}{2}\)

\(=>A\)\(=-\dfrac{3}{2}\left(-\dfrac{3}{2}+3\right)=-2,25\)

Vậy Min A\(=-2,25\)

b,\(B=-x^2-4x-9y^2-6y-6\)

\(=-\left(x^2+4x+4\right)-\left(3y\right)^2-2.3y-1-1\)

\(=-\left(x+2\right)^2-\left(3y+1\right)^2-1\le-1\)

dấu"=' xảy ra\(< =>x=-2,y=-\dfrac{1}{3}\)

a.

$(x+1)(x+2)(x+4)(x+5)=(x+1)(x+5)(x+2)(x+4)=(x^2+6x+5)(x^2+6x+8)$

$=a(a+3)$ với $a=x^2+6x+5$

$=a^2+3a=(a^2+3a+\frac{9}{4})-\frac{9}{4}$

$=(a+\frac{3}{2})^2-\frac{9}{4}$

$=(x^2+6x+\frac{13}{2})^2-\frac{9}{4}\geq \frac{-9}{4}$

Vậy gtnn của biểu thức là $\frac{-9}{4}$. Giá trị này đạt tại $x^2+6x+\frac{13}{2}=0$

$\Leftrightarrow x=\frac{-6\pm \sqrt{10}}{2}$

4M = 4x^2+4y^2-4xy+8x-16y-8072

= [(4x^2-4xy+y^2)-2.(2x+y).2+4]+(3y^2-12y+12)-8088

= [(2x-y)^2-2.(2x-y).2+4]+3.(y^2-4y+4)-8088

= (2x-y-2)^2+3.(y-2)^2-8088 >= -8088

=> M >= -2022

Dấu "=" xảy ra <=> 2x-y-2=0 và y-2=0 <=> x=y=2

Vậy GTNN của M = -2022 <=> x=y=2

Tk mk nha