A = 2x² + 9y² - 6xy - 6x - 12y + 2004

= (x²-6xy+9y²) + 4(x-3y) + 4 + (x²-10x+25) + 1975

= (x-3y)² + 4(x-3y) + 4 + (x-5)² + 1975

= (x-3y+2)² + (x-5)² + 1975 \(\ge\) 1975

Vậy MinA = 1975

Dấu "=" xảy ra khi x = 5; y = \(\dfrac{7}{3}\)

e) 4x² + 15x + 9

= (2x)² + 2x.\(\dfrac{15}{2}\) + \(\dfrac{225}{4}\)-\(\dfrac{225}{4}\) + 9

= (2x+\(\dfrac{15}{2}\))²-\(\dfrac{189}{4}\)

\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}=\dfrac{x+1}{14}+\dfrac{x+1}{15}\)

<=> \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}+\dfrac{x+1}{13}-\dfrac{x+1}{14}-\dfrac{x+1}{15}=0\)

<=> \(\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{15}\right)=0\)

Do: \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}-\dfrac{1}{14}-\dfrac{1}{14}>0\) nên x + 1 = 0

Vậy x = -1

a) DE // AB, DE = \(\dfrac{1}{2}\)AB, IK // AB, IK = \(\dfrac{1}{2}\)AB

=> DE//IK và DE = IK

b) Xét tg GDE và tg GIK có:

DE = IK (cmt)

GDE = GIK (slt)

GED = GKI (slt)

=> tg GDE = tg GIK (g.c.g)

=> GD = GI ( c.t.ứ)

Có GD = GI = IA nên AG = \(\dfrac{2}{3}\)AD

A = \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\left(1-\dfrac{1}{1+2+3+...+2017}\right)\)

= \(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)...\left(1-\dfrac{2}{2017.2018}\right)\)

= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}...\dfrac{2015.2018}{2016.2017}.\dfrac{2016.2019}{2017.2018}\)

= \(\dfrac{2019}{3.2017}=\dfrac{673}{2017}\)

Ta có: \(\dfrac{2016}{2017}=1-\dfrac{1}{2017}\)

\(\dfrac{2017}{2018}=1-\dfrac{1}{2018}\)

Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}\) => \(\dfrac{2016}{2017}< \dfrac{2017}{2018}\)