a) Đa thức M = x² - 2xy + 5x² - 1 = 6x² - 2xy - 1 có bậc 2.

Đa thức N = x²y² - y² + 5x² - 3x²y + 5 có bậc 4.

b) N + M = x2y2 – y2 + 11x2 – 3x2y + 4 – 2xy

Đây ko phải Toán lớp 1 đâu nha anh

a,A=3x^2y^4+5x^3+xy-3x^2y^4

   A=5x³ +xy

=> bậc của A là 3

b,B=7x^3y.(-4x^2y^2)+17x^2y^3-4x^2y+28x^2y^4

  => bậc của B là 8

c,C=5x^4y^2-7x^3y^2.(-2xy^2)-5x^4y^2+x^3-14x^4y^4

   C = 5x⁴y² -7x³y² (-2xy²) - 5x⁴y² +x³ -14x⁴y⁴ 

   C =  5x⁴y² + 14x⁴y⁴ -5x⁴y² +x³ -14x⁴y⁴ 

   C = x³ 

=> Bậc của C là 3

cám ơn

a) Ta có: \(M=x^2y+xy^2-5x^2y^2+x^3-2x^2y+6xy^2\)

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

\(=x^3-x^2y+7xy^2-5x^2y^2\)

Bậc là 4

Ta có: \(N=3x^3+xy+y^2-x^2y^2-2-2xy+7y^2\)

\(=3x^3+\left(xy-2xy\right)+\left(y^2+7y^2\right)-x^2y^2-2\)

\(=3x^2+8y^2-xy-x^2y^2-2\)

Bậc là 4

:/

Bài 3: 

a: \(x^2-16=\left(x-4\right)\cdot\left(x+4\right)\)

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

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

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 )

6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)

7) \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)

8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)

10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)

6) 9x³y² + 3x²y² = 3x²y²( 3x + 1 )

7) x³ + 2x² + 3x = x( x² + 2x + 3 )

8) 6x²y + 4xy² + 2xy = 2xy( 3x + 2y + 1 )

9) 5x²( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )

10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )

a, \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)

b, \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)

c, \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)

d, \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\)

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

Đặt \(x^2=a;y^2=b\left(\text{a,b }\ge0\right)\text{ ta có:}\)

\(a+b=2\)

\(\Rightarrow3a^2+5ab+2b^2+2b\)

\(=\left(3a^2+3ab\right)+\left(2ab+2b^2\right)+2b\)

\(=3a\left(a+b\right)+2b\left(a+b\right)+2b\)

\(=\left(a+b\right)\left(3a+2b\right)+2b\)

\(\text{Mà }a+b=2\text{ nên:}\)

\(=2\left(3a+2b\right)+2b\)

\(=6\left(a+b\right)=6.2=12\) 

Vậy....