Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0

=> 6x² - 21x - (6x² + x - 90x - 15) - 2010 = 0

=> 6x² - 21x - 6x² + 89x + 15 - 2010 = 0

=> 68x - 1995 = 0

 ? 

b) 2x(x - 2012) - x + 2012 = 0

=> 2x(x - 2012) - (x - 2012) = 0

=> (x - 2012) (2x - 1) = 0

⇔[

⇔[

⇔[

Vậy x = {2012;12 }

Ta có : 3x(2x - 7) - (6x + 1)(x - 15) - 2010 = 0

=> 6x² - 21x - (6x² + x - 90x - 15) - 2010 = 0

=> 6x² - 21x - 6x² + 89x + 15 - 2010 = 0

=> 68x - 1995 = 0

 ? 

b) 2x(x - 2012) - x + 2012 = 0

=> 2x(x - 2012) - (x - 2012) = 0

=> (x - 2012) (2x - 1) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2012=0\\2x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2012\\2x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2012\\x=\frac{1}{2}\end{cases}}\)

Vậy x = \(\left\{2012;\frac{1}{2}\right\}\)

a) x²+y²-4x+4y+8=0

⇔ (x-2)²+(y+2)²=0

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

b)5x²-4xy+y²=0

⇔ x²+(2x-y)²=0

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

c)x²+2y²+z²-2xy-2y-4z+5=0

⇔ (x-y)²+(y-1)²+(z-2)²=0

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

b: Ta có: \(5x^2-4xy+y^2=0\)

\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)

\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

d)3x²+3y²+3xy-3x+3y+3=0

⇔ 6x²+6y²+6xy-6x+6y+6=0

⇔ 3(x+y)²+3(x-1)²+3(y+1)²=0

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

1. Ta có: 

\(x^3-9x^2+27x-26=x^3-2x^2-7x^2+14x+13x-26\)

\(=x^2\left(x-2\right)-7x\left(x-2\right)+13\left(x-2\right)=\left(x-2\right)\left(x^2-7x+13\right)\)

Thay x = 23, ta có: \(C=\left(23-2\right)\left(23^2-7.23+13\right)=8001\)

2.

a) \(x^2+4y^2+6x-12y+18=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(4y^2-12y+9\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(2y-3\right)^2=0\)

Mà \(\left(x-3\right)^2\ge0\) với mọi x, \(\left(2y-3\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)và \(\left(2y-3\right)^2=0\Leftrightarrow2y-3=0\Leftrightarrow y=\frac{3}{2}\)

Vậy \(\left(x,y\right)=\left(3;\frac{3}{2}\right)\)

b) \(2x^2+2y^2+2xy-10x-8y+41=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2-10x+25\right)+\left(y^2-8y+16\right)=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-5\right)^2+\left(y-4\right)^2=0\)

.....................................

Rồi giải tương tự như trên

rgthaegƯ mk chỉ giải được phần a thui 

  x^2 + 2y^2 - 2xy + 2x + 2 - 4y =0 
<=>x^2 + y^2 - 2xy+2x-2y+y^2-2y+1+1=0 
<=>(x-y)^2+2(x-y)+1+(y-1)^2=0 
<=>(x-y+1)^2+(y-1)^2=0 
<=>y=1;x=0

Mình làm câu đầu tượng trưng thui nhé, 2 câu sau tương tự vậy !!!!!!

a) pt <=> \(x^2-2xy+2y^2-2x-2y+5=0\)

<=> \(\left(x-y-1\right)^2+y^2-4y+4=0\)

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

TA LUÔN CÓ: \(\left(x-y-1\right)^2;\left(y-2\right)^2\ge0\forall x;y\)

=> \(\left(x-y-1\right)^2+\left(y-2\right)^2\ge0\)      (2)

TỪ (1) VÀ (2) => DẤU "=" SẼ PHẢI XẢY RA <=> \(\hept{\begin{cases}\left(x-y-1\right)^2=0\\\left(y-2\right)^2=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\y=2\end{cases}}\)

VẬY \(\left(x;y\right)=\left(3;2\right)\)