\(-5x^2-2xy-2y^2+14x+10y-1\\ =-\left(x^2+2xy+y^2\right)-\left(4x^2-2\cdot2\cdot\dfrac{7}{2}x+\dfrac{49}{4}\right)-\left(y^2-10y+25\right)+\dfrac{55}{4}\\ =-\left(x+y\right)^2-\left(2x-\dfrac{7}{2}\right)^2-\left(y-5\right)^2+\dfrac{55}{4}\le\dfrac{55}{4}\\ Max\Leftrightarrow\left\{{}\begin{matrix}x=-y\\2x=\dfrac{7}{2}\\y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=\dfrac{7}{4}\\y=5\end{matrix}\right.\Leftrightarrow x,y\in\varnothing\)

Vậy dấu \("="\) ko xảy ra

a: Ta có: \(-x^2+3x\)

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

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

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

a: \(x^2+3y^2-4x+6y+7=0\)

\(\Leftrightarrow x^2-4x+4+3y^2+6y+3=0\)

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

\(\Leftrightarrow\left(x,y\right)=\left(-2;1\right)\)

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

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

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

a) 10x(x-y) - 6y(y-x)
= 10x(x-y) +6y ( x-y)
=(10x+6y) (x-y)
b) 3x² + 5y - 3xy -5x
= 3x(x-y) + 5(y-x)
= 3x(x-y) -5(x-y)
= (3x-5) ( x-y)
c) 3y^₂ - 3z² +3x² + 6xy
=3(y² - z² + x² + 2xy)
=3[(x² +2xy+y²)-z² ]
=3[(x+y)² - z² ]
=3(x+y-z) (x+y+z)
d) 16x³ + 54y3
=2(8x³ + 27y³ )
=2[(2x)³ + (3y)³ ]
=2(2x+3y) (4x² - 6xy + 9y² )
e) x² - 25 -2xy+y²
=(x²-2xy+y²)-25
=(x-y)² -5²
=(x-y-5) (x-y+5)
f) (mình chưa làm ra )
{mong m.n bổ sung thêm..}

mấy câu trên bạn kia đã trả lời rồi nên mk k làm lại nx

f, x⁵ - 3x⁴ + 3x³ - x²

= x² (x³ - 3x² + 3x -1)

= x² (x - 1)³

Chúc bạn học tốt!

x² - 3y² + 2xy + 2x - 4y - 7 = 0

<=> 4.(x² - 3y² + 2xy + 2x - 4y - 7) = 0

<=> 4x² - 12y² + 8xy + 8x - 16y - 28 = 0

<=> (4x² + 8xy + 4y²) + (8x + 8y) + 4 - 16y² - 24y - 32 = 0

<=> (2x + 2y)² + 4(2x + 2y) + 4 - (16y² + 24y + 9) = 23

<=> (2x + 2y + 2)² - (4y + 3)² = 23

<=> (2x + 6y + 5)(2x - 2y - 1) = 23

Vì \(x;y\inℤ\Rightarrow2x+6y+5;2x-2y-1\inℤ\) 

Lập bảng : 

Vậy (x;y) = (3;2) ; (-9;2)