A=|2x-2019|+|2x-3|

A=|2x-2019|+|3-2x|          (Vì |A|=|-A| nha bạn)

A lớn hơn hoặc =|2x-2019+3-2x|

                          =2016

Vậy GTNN A=2016

cảm ơn bạn nhiều, bạn cứu mình rồi

\(4B=4x^2+4xy+4y^2-8x-12y+8076\)

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

\(+\left(2x\right)^2-8x+8076\)

= \(\left(2y-3+x\right)^2+3x^2-2x+8076\)

đến đây thì dễ rồi

1. B = | x - 2018 | + | x - 2019 | + | x - 2020 |

= ( | x - 2018 | + | x - 2020 | ) + | x - 2019 | 

= ( | x - 2018 | + | 2020 - x | ) + | x - 2019 |

Vì \(\hept{\begin{cases}\left|x-2018\right|+\left|2020-x\right|\ge\left|x-2018+2020-x\right|=2\\\left|x-2019\right|\ge0\end{cases}}\)=> B ≥ 2 ∀ x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left(x-2018\right)\left(2020-x\right)\ge0\\x-2019=0\end{cases}}\Rightarrow x=2019\)

Vậy MinB = 2 <=> x = 2019

2. ĐKXĐ : x ≥ 0

Ta có : \(\sqrt{x}+3\ge3\forall x\ge0\)

=> \(\frac{2019}{\sqrt{x}+3}\le673\forall x\ge0\). Dấu "=" xảy ra <=> x = 0 (tm)

Vậy MaxC = 673 <=> x = 0

Bài 1 : 

\(B=\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\)

Ta có : \(\left|x-2018\right|\ge0\forall x;\left|x-2019\right|\ge0\forall x;\left|x-2020\right|\ge0\forall x\)

\(\left|x-2018\right|+\left|x-2019\right|+\left|x-2020\right|\ge0\)

Dấu ''='' xảy ra khi \(x=2018;x=2019;x=2020\)

Vậy GTNN B là 0 khi x = 2018 ; x = 2019 ; x = 2020 

\(\frac{2\left|2018x-2019\right|+2019}{\left|2018x-2019\right|+1}\)

\(=\frac{\left(2\left(\left|2018x-2019\right|+1\right)\right)+2017}{\left|2018x-2019\right|+1}\)

\(=2+\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất

\(\Rightarrow\frac{2017}{\left|2018x-2019\right|+1}\)có giá trị lớn nhất

\(\Rightarrow\left|2018x-2019\right|+1\)có giá trị nhỏ nhất

Mà \(\left|2018x-2019\right|\ge0\)

\(\Rightarrow\left|2018x-2019\right|+1\ge1\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left|2018x-2019\right|=0\)

\(\Leftrightarrow x=\frac{2019}{2018}\)

Vậy \(M_{MAX}=2019\)tại \(x=\frac{2019}{2018}\)

\(\frac{5^x+5^{x+1}+5^{x+2}}{31}=\frac{3^{2x}+3^{2x+1}+3^{2x+2}}{13}\)

\(\Rightarrow\frac{5^x\left(1+5+5^2\right)}{31}=\frac{3^{2x}\left(1+3+3^2\right)}{13}\)

\(\Rightarrow\frac{5^x\cdot31}{31}=\frac{3^{2x}\cdot13}{13}\)

\(\Rightarrow5^x=3^{2x}\)

Mà \(\left(5;3\right)=1\)

\(\Rightarrow x=2x=0\)

trả lời...............................

ok..................................

hk tốt...............................

\(A=2.\left|x-\frac{1}{2}\right|-2019\)

Vì \(\left|x-\frac{1}{2}\right|\ge0,\forall x\)

\(\Rightarrow2.\left|x-\frac{1}{2}\right|\ge0,\forall x\)

\(\Rightarrow2.\left|x-\frac{1}{2}\right|-2019\ge-2019,\forall x\)

Dấu \("="\)xảy ra 

\(\Leftrightarrow\left|x-\frac{1}{2}\right|=0\)

\(x-\frac{1}{2}=0\)

\(x=0+\frac{1}{2}\)

\(x=\frac{1}{2}\)

Vậy \(A_{min}=-2019\Leftrightarrow x=\frac{1}{2}\)

 \(A=2.\left|x-\frac{1}{2}\right|-2019\)

Ta có : \(2.\left|x-\frac{1}{2}\right|\ge0\forall x\)

\(\Rightarrow2\left|x-\frac{1}{2}\right|-2019\ge-2019\)

Dấu "=" xảy ra \(\Leftrightarrow2.\left|x-\frac{1}{2}\right|=0\Leftrightarrow x-\frac{1}{2}=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy : \(A_{min}=-2019\) tại \(x=\frac{1}{2}\)

\(B=4\left|3x-2\right|+3\left|4x+1\right|-\frac{1}{3}\)

Ta có : \(4\left|3x-2\right|\ge0\forall x,3\left|4x+1\right|\ge0\forall x\)

\(\Rightarrow4\left|3x-2\right|+3\left|4x+1\right|\ge0\forall x\)

\(\Rightarrow4\left|3x-2\right|+3\left|4x+1\right|-\frac{1}{3}\ge-\frac{1}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-2=0\\4x+1=0\end{cases}}\)

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