ĐK: x khác 0

Từ\(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)

\(\Rightarrow x^2+2+\frac{1}{x^2}+x^2+xy+\frac{y^2}{4}=6+xy\)

\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+\left(x+\frac{y}{2}\right)^2=6+xy\)

Do VT > 0\(\Rightarrow6+xy\ge0\Rightarrow xy\ge6\)
Có A = 2016 + xy > 2016 + 6 = 2022

tth : Viết nhầm :V
Đoạn cuối \(6+xy\ge0\Rightarrow xy\ge-6\)

Có A = 2016 + xy > 2016 - 6 = 2010 !!!

Được rồi chứ gì -.- 

\(x^3+3xy+y^3\)

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

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

\(=x^2+2xy+y^2\)

\(=\left(x+y\right)^2\)

\(=1^2\)

\(=1\)

\(x^3+3xy+y^3=x^3+3xy.1+y^3\)

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

                             \(=x^3+3x^2y+3xy^2+y^3\)

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

\(x.\left(x-2009\right)-2010x+2009.2010=0\)

\(x.\left(x-2009\right)-2010\left(x-2009\right)=0\)

\(\left(x-2009\right)\left(x-2010\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2009=0\\x-2010=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2009\\x=2010\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=2009\\x=2010\end{cases}}\)

dat a =2009-x

b=x-2010

ta co : a^2+ab+b^2/a^2-ab+b^2 =19/49

<=>49a^2+49ab+49b^2=19a^2-19a+19b^2

<=>30a^2+68a+30b^2=0

<=>15a^2+34ab+15b^2=0

<=>15a^2+9ab+25ab+15b^2=0

<=>3a(5a+3b)+5b(5a+3b)=0

<=>(5a+3b)(3a+5b)=0

<=>5a+3b=0 hoac 3a+5b=0

vs 5a +3b=0 <=>5(2009-x)+3(x-2010)=0=>x=......

x^2-3x=0

\(\Leftrightarrow\)x(x-3)=0

\(\Leftrightarrow\)X=0 hay x-3=0

\(\Leftrightarrow\)x=0 hay x=3

\(x^2-3x=0\)

\(x\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

\(B=\frac{x^2-6x+14}{x^2-6x+12}=\frac{x^2-6x+12+2}{x^2-6x+12}=1+\frac{2}{x^2-6x+12}\)

ta có: \(x^2-6x+12=x^2-2.3.x+3^2+4=\left(x-3\right)^2+4\ge4\)

để Bmax => \(\left(\frac{2}{x^2-6x+12}\right)max\Rightarrow x^2-6x+12min\)và lớn hơn 0 vì 2>0

mà \(\left(x-3\right)^2+4\) \(\ge\)4

dấu = xảy ra khi x-3=0

=> x=3

Vậy \(MaxB=\frac{3}{2}\)khi x=3

lùa đảo à

bạn k lm đc thì thôi đừng nói nhiều nha