a, Vì |x-3| \(\ge\)0

=>A=|x-3|+50\(\ge\)50

Dấu "=" xảy ra khi x=3

Vậy GTNN của A = 50 khi x=3

b, Vì |x+8| \(\ge0\)

=>B=2014-|x+8|\(\le2014\)

Dấu "=" xảy ra khi x=-8

Vậy GTLN của B = 2014 khi x=-8

c, Vì \(\hept{\begin{cases}\left|x-100\right|\ge0\\\left|y+2014\right|\ge0\end{cases}}\)

\(\Rightarrow\left|x-100\right|+\left|y+2014\right|\ge0\)

\(\Rightarrow C=\left|x-100\right|+\left|y+2014\right|-2015\ge-2015\)

Dấu "=" xảy ra khi x=100,y=-2014

Vậy GTNN của C=-2015 khi x=100,y=-2014

\(x = {{b^2} \over 2a}\)

a)4x(x+1)+(3-2x)=15

  4x² + 4x + 3 - 2x = 15

  4x² + 2x -12 = 0

  4x² + 8x - 6x - 12 = 0

  4x(x+2) - 6(x+2) = 0

   (x+2)(4x - 6)= 0 => TH1: x = -2 . TH2: x = 3/2

b)9x(x-2014)-x+2014=0

9x(x-2014)-(x-2014) = 0

(x-2014) (9x -1) = 0 => TH1: x = 2014. TH2: x=1/9

a)x=4

còn b mk thấy vô lý quá

1)

\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)

\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)

\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)

\(\Leftrightarrow x=2015\)

Vậy \(S=\left\{2015\right\}\)

<

\(x+\frac{5}{2015}+x+\frac{6}{2014}+x+\frac{3}{2017}=-3\)

\(\Rightarrow3x+\left(\frac{1}{403}+\frac{3}{1007}+\frac{3}{2017}\right)=-3\)

\(\Rightarrow\frac{1}{403}+\frac{3}{1007}+\frac{3}{2017}=-3-3x=-3.\left(1-x\right)\)

\(\Rightarrow\frac{\frac{1}{403}+\frac{3}{1007}+\frac{3}{2017}}{-3}-1=-x\)

Cảm ơn nha

vì (2x-1)^2014 + (y-2/5)^2014 + /x+y-z/=0

(2x-1)^2014=0

((y-2/5)^2014=0

/x+y+z/=0

vậy 2x-1=0 thì x=1/2

y-2/5=0 thì y=2/5

x+y+z=0=1/2 +2/5 +z=0 thi z=-9/10

mk mới học lớp 5 thôi

Ta có:

\(16\times17\times18\times19\times20=1860a80.\)

Mà \(16\times17\times18\times19\times20=1860480\)

\(\Rightarrow1860a80=1860480\)

Ta thấy a = 4

Vậy a = 4

\(a=4\)

\(HT\)