\(2x^2+y^2+2x-2xy+5-4y=0\)

\(\Leftrightarrow\left[y^2-2y\left(x+2\right)+\left(x+2\right)^2\right]+\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow\left(y-x-2\right)^2+\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y-x-2=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)

\(S=\left(x+2\right)^2+\left(y-1\right)^2=\left(1+2\right)^2+\left(3-1\right)^2\)

\(=3^2+2^2=13\)

\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)

bn còn ngu hơn bn thảo

= 2016x - 1 = 1

x = 2/2016 = 1/ 1008

ngu tkees

\(\left(2016x-1\right)^{2016}=1^{2016}\)\(\Rightarrow2016x-1=1\Rightarrow2016x=0\Rightarrow x=2016\)

a ) 8x - 6x - 15 = 13

      2x = 13 + 15

       2x = 28

            x = 28 / 2   => x = 14

b) 19 + 25 = x + x + 2x -4

 19 + 25 + 4 = 4x

 48 = 4x

=> x = 48/4 = 12

c ) 49 - 14 = 6x - x + 5

  49 - 14 -5 = 5x

  30 = 5x

x = 30/5

=> x = 6

8x - 6x - 15 = 13

=> 2x = 28

=> x = 14

19 + 25 = x + x + 2x - 4

=> 44 = 4x - 4

=> 44 + 4 = 4x

=> 48 = 4x

=> x = 12

49 - 14 = 6x - x + 5

=> 35 = 5x + 5

=> 5x = 30

=> x = 6

+)8x - 6x - 15 = 13

2x - 15 = 13

2x = 28

x = 14

+) 19 + 25 = x + x + 2x - 4

44 = 4x - 4

48 = 3x

x =12

+) 49 - 14 = 6x - x + 5

35 = 5x + 5

40 = 5x

x = 8

Tk mk nha! 

\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)

Ta có:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\)

\(=\frac{\left(2x+3y-z\right)-5}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)

\(\Rightarrow\begin{cases}x-1=2.5=10\\y-2=3.5=15\\z-3=4.5=20\end{cases}\)\(\Rightarrow\begin{cases}x=11\\y=17\\z=23\end{cases}\)

Vậy x = 11; y = 17; z = 23