Do A nhỏ nhất 

Suy ra : x^2 = 0, 2y^2 = 0 , 4y = 0 .......( tất cả số hạng bằng 0) 

Suy ra A= 2019

\(A=x^2+2y^2+4y+2xy-4x+2019\)

\(A=\left(x^2+y^2-2^2+2xy-4y-4x\right)+\left(y^2+8y+4^2\right)+2007\)

\(A=\left(x+y-2\right)^2+\left(y+4\right)^2+2007\ge2007\)

Vậy \(Min_A=2007\) khi \(\hept{\begin{cases}x+y-2=0\\y+4=0\end{cases}}\hept{\begin{cases}x+y=2\\y=-4\end{cases}}\hept{\begin{cases}x=6\\y=4\end{cases}}\)

\(\left(x-3\right)\left(4x+5\right)+2019\)

\(=4x^2-7x-15+2019\)

\(=4x^2-7x+2004\)

\(=4\left(x^2-\frac{7}{4}x+501\right)\)

\(=4\left(x^2-\frac{7}{4}x+\frac{49}{64}+\frac{32015}{64}\right)\)

\(=4\left[\left(x-\frac{7}{8}\right)^2+\frac{32015}{64}\right]\)

\(=4\left[\left(x-\frac{7}{8}\right)^2\right]+\frac{32015}{16}\ge\frac{32015}{16}\)

Vậy GTNN của bt là \(\frac{32015}{16}\Leftrightarrow x-\frac{7}{8}=0\Leftrightarrow x=\frac{7}{8}\)

mk tưởng 7/4

a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

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

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

nhanh lên các bạn

1) \(4x^2+4x+3=\left(2x+1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{2021}{4x^2+4x+3}\le\dfrac{2021}{2}\Rightarrow H=-\dfrac{2021}{4x^2+4x+3}\ge-\dfrac{2021}{2}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{1}{2}\)

2) \(5x^2-2x+10=5\left(x^2-\dfrac{2}{5}x+\dfrac{1}{25}\right)+\dfrac{49}{5}=5\left(x-\dfrac{1}{5}\right)^2+\dfrac{49}{5}\ge\dfrac{49}{5}\)

\(\Rightarrow I=\dfrac{-2019}{5x^2-2x+10}\ge-\dfrac{10095}{49}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{5}\)

ghi lộn môn hc ạ 

\(A=\left|4x-3\right|+\left|5y+7,5\right|+17,5\)

Ta thấy \(\left|4x-3\right|\ge0;\left|5y+7,5\right|\ge0\)

\(\Rightarrow\left|4x-3\right|+\left|5y+7,5\right|+17,5\ge17,5\)

\(\Rightarrow A\ge17,5\)

Dấu "=" xảy ra  \(\Leftrightarrow\hept{\begin{cases}4x-3=0\\5y+7,5=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\y=-1,5\end{cases}}\)

...
\(B=\left|x-2\right|+\left|x-6\right|+2017\)

\(=\left|x-2\right|+\left|6-x\right|+2017\)

Ta thấy \(\left|x-2\right|+\left|6-x\right|\ge\left|x-2+6-x\right|=4\)

\(\Rightarrow B\ge4+2017=2021\)

Dấu "=" xảy ra khi \(2\le x\le6\)

....

\(C=\left(2x+1\right)^{2020}-2019\)

Ta thấy \(\left(2x+1\right)^{2020}\ge0\)

\(\Rightarrow C=\left(2x+1\right)^{2020}-2019\ge-2019\)

Dấu "=" xảy ra khi \(2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

....