`Answer:`

\(a)\left(-3x^2y-2xy^2+6\right)+\left(-x^2y+5xy^2-1\right)\)

\(=-3x^2y-2xy^2+6+-x^2y+5xy^2-1\)

\(=\left(-3x^2y-x^2y\right)+\left(-2xy^2+5xy^2\right)+\left(6-1\right)\)

\(=-4x^2y+3xy^2+5\)

\(b)\left(1,6x^3-3,8x^2y\right)+\left(-2,2x^2y-1,6x^3+0,5xy^2\right)\)

\(=1,6x^3-3,8x^2y+-2,2x^2y-1,6x^3+0,5xy^2\)

\(=\left(1,6x^3-1,6x^3\right)+\left(-3,8x^2y+-2,2x^2y\right)+0,5xy^2\)

\(=-6x^2y+0,5xy^2\)

\(c)\left(6,7xy^2-2,7xy+5y^2\right)-\left(1,3xy-3,3xy^2+5y^2\right)\)

\(=6,7xy^2-2,7xy+5y^2-1,3xy+3,3xy^2-5y^2\)

\(=\left(6,7xy^2+3,3xy^2\right)+\left(-2,7xy-1,3xy\right)+\left(5y^2-5y^2\right)\)

\(=10xy^2+-4xy\)

\(=10xy^2-4xy\)

\(d)\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=\left(3x^2+x^2-4x^2\right)+\left(-2xy-xy\right)+\left(y^2+2y^2+y^2\right)\)

\(=-3xy+4y^2\)

\(e)\left(x^2+y^2-2xy\right)-\left(x^2+y^2+2xy\right)+\left(4xy-1\right)\)

\(=x^2+y^2-2xy-x^2-y^2-2xy+4xy-1\)

\(=\left(x^2-x^2\right)+\left(y^2-y^2\right)+\left(-2xy-2xy+4xy\right)-1\)

\(=-1\)

a) \(3x^2-3xy-5x+5y\)

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

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

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

b) \(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy\left(x^2-y^2-2y-1\right)\)

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

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

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

c) \(x^2+1+2x-y^2\)

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

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

\(=\left(x+1+y\right)\left(x+1-y\right)\)

d) \(x^2+4x-2xy-4y+y^2\)

\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)

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

\(=\left(x-y\right)\left(x-y+4\right)\)

e) \(x^3-2x^2+x\)

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

\(=x\left(x-1\right)^2\)

f) \(2x^2+4x+2-2y^2\)

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

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

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

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

a: =3x(x-y)-5(x-y)

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

b: \(=2xy\left(x^2-y^2-2y-1\right)\)

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

\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)

d:

Sửa đề: x^2+4x-2xy-4y+y^2

=x^2-2xy+y^2+4x-4y

=(x-y)^2+4(x-y)

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

e: =x(x^2-2x+1)

=x(x-1)^2

f: =2(x^2+2x+1-y^2)

=2[(x+1)^2-y^2]

=2(x+1+y)(x+1-y)

a) \(x^2-3xy+3y^2=3y\)

Rõ ràng \(x⋮y\) nên đặt \(x=ky\left(k\inℤ\right)\). Pt trở thành:

\(k^2y^2-3ky^2+3y^2=3y\)

\(\Leftrightarrow\left[{}\begin{matrix}y=0\\k^2y-3ky+3y=3\end{matrix}\right.\).

Khi \(y=0\) \(\Rightarrow x=0\).

Khi \(k^2y-3ky+3y=3\)

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

Ta lập bảng giá trị:

Vậy pt đã cho có các nghiệm \(\left(0;0\right);\left(3;1\right);\left(3;3\right);\left(6;3\right)\)

b) \(x^2-2xy+5y^2=y+1\)

\(\Leftrightarrow x^2-2yx+5y^2-y-1=0\)

\(\Delta'=\left(-y\right)^2-\left(5y^2-y-1\right)\) \(=-4y^2+y+1\)

Để pt đã cho có nghiệm thì \(-4y^2+y+1\ge0\), giải bpt thu được \(\dfrac{1-\sqrt{17}}{8}\le y\le\dfrac{1+\sqrt{17}}{8}\). Mà lại có \(-1< \dfrac{1-\sqrt{17}}{8}< 0< \dfrac{1+\sqrt{17}}{8}< 1\) nên suy ra \(y=0\). Từ đó tìm được \(x=\pm1\). Vậy pt đã cho có các nghiệm \(\left(1;0\right);\left(-1;0\right)\)

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

sửa lại:x=1

1/x^3 - 2x^2 - 9x + 18

= x\(^2\)( x - 2 ) - 9 ( x - 2 ) = ( x\(^2\) - 9 ) ( x - 2 )= ( x - 3 ) ( x +3 ) ( x - 2 )

2/3x^2 -5x - 3y^2 + 5y

= 3( x\(^2\) - y\(^2\) ) - 5 ( x - y ) = 3 ( x - y ) ( x + y ) - 5 ( x - y ) = ( x - y ) [ 3( x+ y ) - 5 ]

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

3/49 - x^2 + 2xy - y^2

= 49 - ( x\(^2\) - 2xy + y\(^2\) ) = 49 - ( x - y )\(^2\) = ( 7 - x + y ) ( 7 + x - y )

5/ x^2 - 4x^2y^2 + 2xy

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

6/ 3x - 3y - x^2 + 2xy - y^2

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

Lâu rồi  hổng thấy ai giải nên giải luôn ak 

Ta có \(5x^2+2xy+2y^2=\left(2x+y\right)^2+\left(x-y\right)^2\ge\left(2x+y\right)^2\Rightarrow\sqrt{5x^2+2xy+2y^2}\ge2x+y.\)

           \(2x^2+2xy+5y^2=\left(x+2y\right)^2+\left(x-y\right)^2\ge\left(x+2y\right)^2\Rightarrow\sqrt{2x^2+2xy+5y^2}\ge x+2y.\)

Suy ra \(Q\ge3\left(x+y\right)=3.1=3\)dấu = xảy ra khi \(\hept{\begin{cases}x+y=1\\x-y=0\end{cases}\Leftrightarrow}x=y=\frac{1}{2}\)

b)x²+2xy+y²-16=(x+y)²-4²=(x+y+4)(x+y-4)

c)3x²+5x-3xy-5y=x(3x+5)-y(3x+5)=(3x+5)(x-y)

d)4x²-6x³y-2x²+8x=2x(2x-3x²y-x+4)

e)x²-4-2xy+y²=(x²-2xy+y²)-4=(x-y)²-2²=(x-y-2)(x-y+2)

k)x²-y²-z²-2yz=x²-(y+z)²=(x-y-z)(x+y+z)

m)6xy+5x-5y-3x²-3y²=3(x²-2xy+y²)+5(x-y)=3(x-y)²+5(x-y)=(x-y)(3x-3y+5)

b. (x^2+2xy+y^2)-16 =(x+y)^2-16=(x+y+4)(x+y-4)