\(a.A=5xy^2+xy-3xy^2-x^2y+2xy+x^2y-2xy^2+xy+4\\ =\left(5xy^2-3xy^2-2xy^2\right)+\left(xy+2xy+xy\right)+\left(-x^2y+x^2y\right)+4\\ =4xy+4\)

Bậc của A là: 2 

b. Thay `x=2;y=1` vào A ta có:  

\(A=4\cdot2\cdot1+4=12\) 

\(c.A+B=-2xy+1\\ =>B=-2xy+1-A\\ =>B=\left(-2xy+1\right)-\left(4xy+4\right)\\ =-2xy+1-4xy-4\\ =-6xy-3\)

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

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

`= 0 + 0 + 4xy + 4`

`= 4xy + 4`

Bậc: 2

b) Thay `x = 2; y = 1` vào `A` ta được: 

`A = 4 . 2 . 1 + 4 = 8 + 4 = 12`

c) Ta có: `A + B = -2xy + 1`

`=> B =  -2xy + 1 - A`

`=> B =  -2xy + 1 - (4xy + 4)`

`=> B =  -2xy + 1 - 4xy - 4`

`=> B =  -6xy - 3`

Vậy ....

g: \(\left(x+3\right)^2-\left(x-4\right)\left(x+4\right)=1\)

=>\(x^2+6x+9-x^2+16=1\)

=>6x=1-9-16=-8-16=-24

=>x=-4

i: \(\left(x-5\right)^2-\left(x-3\right)\left(x+3\right)=1\)

=>\(x^2-10x+25-\left(x^2-9\right)=1\)

=>\(x^2-10x+25-x^2+9=1\)

=>-10x=1-25-9=-24-9=-33

=>\(x=\dfrac{33}{10}\)

j: \(3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\)

=>\(3\left(x^2+4x+4\right)+4x^2-4x+1-7\left(x^2-9\right)=36\)

=>\(3x^2+12x+12+4x^2-4x+1-7x^2+63=36\)

=>8x+76=36

=>8x=-40

=>x=-5

o: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-19\)

=>\(6x^2+2-6x^2+12x-6=-19\)

=>12x-4=-19

=>12x=-15

=>\(x=-\dfrac{15}{12}=-\dfrac{5}{4}\)

Áp dụng định lý Pythagore cho tam giác ABC vuông tại A ta có: 

\(AB^2+AC^2=BC^2\\ =>BC=\sqrt{AB^2+AC^2}=\sqrt{3^2+4^2}=5\left(cm\right)\)

Áp dụng hệ thức lượng ta có:

\(AB^2=BC\cdot BH=>BH=\dfrac{AB^2}{BC}=\dfrac{3^2}{5}=\dfrac{9}{5}\left(cm\right)\) 

\(AB\cdot AC=AH\cdot BC=>AH=\dfrac{AB\cdot AC}{BC}=\dfrac{3\cdot4}{5}=\dfrac{12}{5}\left(cm\right)\)

\(=>S_{HAB}=\dfrac{1}{2}\cdot AH\cdot BH=\dfrac{1}{2}\cdot\dfrac{12}{5}\cdot\dfrac{9}{5}=\dfrac{54}{25}\left(cm^2\right)\)

\(\dfrac{1}{3}\left(6xy^2\right)^2\cdot\left(-\dfrac{5}{4}x^4y^3\right)\\ =\dfrac{1}{3}\cdot36x^2y^4\cdot\left(-\dfrac{5}{4}x^4y^3\right)\\ =\left(\dfrac{1}{3}\cdot36\cdot\dfrac{-5}{4}\right)\cdot\left(x^2\cdot x^4\right)\cdot\left(y^4\cdot y^3\right)\\ =-15x^6y^7\)

Bậc: `6+7=13` 

\(\dfrac{1}{3}\left(6xy^2\right)^2\cdot\left(-\dfrac{5}{4}x^4y^3\right)=\dfrac{1}{3}\cdot36x^2y^4\cdot\dfrac{-5}{4}x^4y^3\)

\(=12\cdot\dfrac{-5}{4}\cdot x^6y^7=-15x^6y^7\)

Bậc là 6+7=13

\(\left(3x-5\right)^2-2x\left(4x-1\right)\)

\(=9x^2-30x+25-8x^2+2x\)

\(=x^2-28x+25\)

\(=x^2-28x+196-171\)

\(=\left(x-14\right)^2-171=\left(x-14-\sqrt{171}\right)\left(x-14+\sqrt{171}\right)\)

\(x^2-2014x+2013\\ =x^2-2013x-x+2013\\ =\left(x^2-2013x\right)-\left(x-2013\right)\\ =x\left(x-2013\right)-\left(x-2013\right)\\ =\left(x-2013\right)\left(x-1\right)\)

\(a.3x^2y-12xy^2\\ =3xy\cdot x-4y\cdot3xy\\ =3xy\left(x-4y\right)\\ b.x^2y-8xy\\ =xy\cdot x-8\cdot xy\\ =xy\left(x-8\right)\)