10x=6y

=> \(\frac{x}{6}=\frac{y}{10}\)

= \(\frac{x^2}{36}=\frac{y^2}{100}\)

= \(\frac{2x^2}{72}=\frac{y^2}{100}=\frac{2x^2-y^2}{72-100}=\frac{16}{-28}=\frac{-4}{7}\)

=> x= -4/7.6= -24/7

=> y= -4/7.10=-40/7

10x = 6y <=> 5x = 3y \(\Rightarrow\frac{x}{3}=\frac{y}{5}\) 

\(\Rightarrow\frac{2x^2}{18}=\frac{y^2}{25}=\frac{2x^2-y^2}{18-25}=\frac{16}{-7}=\)...

(số hơi lẻ)

\(x^2-y^2+10x-6y+16\)

\(=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)

\(=\left(x+5\right)^2-\left(y+3\right)^2\)

\(=\left(x+5-y-3\right)\left(x+5+y+3\right)\)

\(=\left(x-y+2\right)\left(x+y+8\right)\)

:))

\(10x^2+5y^2-2xy-38x-6y+41=0\)

\(\Leftrightarrow\left[\left(x-y\right)^2-2\left(x-y\right)+1\right]+\left(9x^2-36x+36\right)+\left(4y^2-6y+4\right)=0\)

\(\Leftrightarrow\left(x-y-1\right)^2+\left(3x-6\right)^2+\left(2y-2\right)^2=0\)

\(\Leftrightarrow x=2;y=1\)

Sao tìm luôn được nghiệm nhỉ :V chả nhẽ phương trình ( 2 ) chỉ để thử nghiệm thôi sao ?

Điều kiện \(\hept{\begin{cases}x^3+xy+6y\ge0\\y^3+x^2-1\ge0\end{cases}}\)

Ta có pt (1) \(\Leftrightarrow10x^2-2x\left(y+19\right)+5y^2-6y+41=0\)

Tính \(\Delta'_x=-49\left(y-1\right)^2\ge0\Leftrightarrow y\ge1\)thay vào (1) ta được x=2 thỏa mãn hệ phương trình

KL: S={(2;1)}

\(x^2+3y^2-4x+6y+7=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(3y^2+6y+3\right)=0\\ \Leftrightarrow\left(x-2\right)^2+3\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

\(3x^2+y^2+10x-2xy+26=0\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(2x^2+10x+\dfrac{25}{8}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x^2+2\cdot\dfrac{5}{2}x+\dfrac{25}{4}\right)+\dfrac{183}{8}=0\\ \Leftrightarrow\left(x-y\right)^2+2\left(x+\dfrac{5}{2}\right)^2+\dfrac{183}{8}=0\\ \Leftrightarrow x,y\in\varnothing\)

Sửa đề: \(3x^2+6y^2-12x-20y+40=0\)

\(\Leftrightarrow\left(3x^2-12x+12\right)+\left(6y^2-20y+\dfrac{50}{3}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y^2-2\cdot\dfrac{5}{3}y+\dfrac{25}{9}\right)+\dfrac{34}{3}=0\\ \Leftrightarrow3\left(x-2\right)^2+6\left(y-\dfrac{5}{3}\right)^2+\dfrac{34}{3}=0\\ \Leftrightarrow x,y\in\varnothing\)

\(2\left(x^2+y^2\right)=\left(x+y\right)^2\\ \Leftrightarrow2x^2+2y^2=x^2+2xy+y^2\\ \Leftrightarrow x^2-2xy+y^2=0\\ \Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x-y=0\Leftrightarrow x=y\)

xy là x.y hay là x và y vậy bn

X và y là số nguyên phải ko

 x^2-y^2+10x-6y+16=(x^2-10x+25)-(y^2+6y+9)=(x-5)^2-(y+3)^2=(x-5-y-3)(x-5+y+3)=(x-y-8)(x+y-2)

  (x²-10x+25)-(y²+6y+9)=(x-5)²-(y+3)²=(x-y-8)(x+y-2)

\(x^2-y^2+10x-6y+16=\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)(tách 16 thành 25 - 9 xog nhóm vào)

\(=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5+y+3\right).\left(x+5-y-3\right)\)

\(=\left(x+y+8\right).\left(x-y+2\right)\)

Chúc bạn học tốt . 

  x^2−y^2+10x−6y+16
=(x^2+10x+25)-(y^2+6y+9)

=(x+5)^2-(y+3)^2

=(x+5+y+3)(x+5-y-3)

=(x+y+8)(x-y+2)

phân tích đa thức thành nhân tử

a, 6x^2 + 7xy + 2y^2

=6x^2+3xy+4xy+2y^2

=3x(x+y)+2y(x+y)

=(3x+2y)(x+y)

b, 9x^2 - 9xy - 4y^2

=9x^2 +3xy-12xy-4y^2

=3x(x+y)-4y(x+y)

=(3x+4y)(x+y)

c, x^2 - y^2 + 10x - 6y + 16=x^2-y^2+6x-6y+4x+16=x(x+6)-y(x+6)+4(x+6)=(x-y+4)(x+6)

Bài làm

a, 6x² + 7xy + 2y²

= 6x² + 3xy + 4xy + 2y² 

= ( 6x² + 3xy ) + ( 4xy + 2y² )

= 3x( 2x + y ) + 2y( 2x + y )

= ( 2x + y )( 3x + 2y )

b, 9x² - 9xy - 4y² 

= 9x² - 12xy + 3xy - 4y² 

= ( 9x² - 12xy ) + ( 3xy - 4y² )

= 3x( 3x - 4y ) + y ( 3x - 4y )

= ( 3x + y )( 3x - 4y )

c, x² - y² + 10x - 6y + 16

 = x² - y² - 6x + 6y + 4x + 16

= x( x + 6 ) - y( x + 6 ) + 4( x + 6 )

= ( x - y + 4 )( x + 6 )

# Học tốt #