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

a/ x² – 5y + y² -2xy + 5x = ( x² - 2xy + y² ) - 5( y - x ) = ( x - y )² - 5( y - x ) = ( y - x )² - 5( y - x ) = ( y - x )( y - x - 5  )

b/ 4x² – 81(y – 2)² = 4x²  - 9²(y – 2)²=  4x² – ( 9y – 18)² = ( 2x -9y -18 )( 2x + 9y + 18 )

c/ x²z – y²z + 2yz – z = ( x²z + yz ) - ( y²z - yz ) - z = z( x² + y ) - z( y² - y ) -z = z( x² + y - y² +y - 1 ) = z( x² + 2y - y² - 1 ) \(=z[x^2-\left(y^2-2y+1\right)]=z[x^2-\left(y-1\right)^2=z\left(x-y+1\right)\left(x+y-1\right)\)

d/ x³ – 8y³ + x² + 2xy + 4y² = ( x³ – 8y³ ) + x² + 2xy + 4y² = ( x -2y )( x² + 2xy + 4y² ) +  ( x² + 2xy + 4y² 0 = ( x² + 2xy + 4y²)( x -2y +1)

e/ 7x² – 11x + 4 =  7x² -7x -4x +4 = 7x( x-1 ) - 4( x - 1 ) = ( x - 1 )( 7x - 4 )

g/ 13x² + 2xy – 15y² = 13x² - 13xy + 15xy - 15y² = 13x( x - y ) + 15y( x - y ) = ( x - y )( 13x + 15y )

h/ x³ + 3x² + 3x + 2 =  x³ +2x² + x² +2x + x + 2 = x²( x + 2 ) + x( x + 2 ) + ( x + 2 ) = ( x + 2 )( x² + x + 1 )

i/ x³ – 3x² + 3x – 2 + xy – 2y = x³ - 2x² - x² + 2x + x - 2 +xy - 2y = x²( x - 2 ) - x( x - 2 ) + ( x - 2 ) + y( x - 2 ) = ( x - 2 )( x² - x +1 + y )

\(\left(2x-5\right)\left(3x+1\right)-6x\left(x-3\right)\)

\(=6x^2+2x-15x-5-6x^2+18x=5x-5\)

`(2x-5)(3x+1)-6x (x-3)`

`= 6x^2 +2x-15x-5 - 6x^2 + 18x`

`= (6x^2 - 6x^2)+(2x-15x+18x)-5`

`= 5x-5`

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

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

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

ta có 

\(A+B+C+D=360^0\text{ nên }C+D=360^0-A-B\)

bài 1. ta sẽ có : \(\hept{\begin{cases}C+D=360^0-50^0-50^0=260^0\\C=3D\end{cases}\Leftrightarrow\hept{\begin{cases}C=195^0\\D=65^0\end{cases}}}\)

bài 2. ta sẽ có : \(\hept{\begin{cases}C+D=360^0-50^0-70^0=240^0\\C=2D\end{cases}\Leftrightarrow\hept{\begin{cases}C=160^0\\D=80^0\end{cases}}}\)

bài 1. ta sẽ có : \(\hept{\begin{cases}C+D=360^0-60^0-60^0=240^0\\C-D=20^0\end{cases}\Leftrightarrow\hept{\begin{cases}C=130^0\\D=110^0\end{cases}}}\)

\(8x\left(x-2\right)-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-16x-3x^2+12x+15-5x^2\)

\(=15-4x\)

`8x(x-2) -3 (x^2 -4x-5)-5x^2`

`= 8x^2 - 16x - 3x^2 +12x+15 - 5x^2`

`= (8x^2 - 3x^2 - 5x^2)+(-16x +12x)+15`

`= -4x +15`

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

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

LÀM TIẾP NHA

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

\(\left(-2a+3\right)\left(-2+3a\right)=4a-6a^2-6+9a=6a^2+13a-6\)

`(-2a+3) (-2+3a)`

`= -2a (-2+3a)+3 (-2+3a)`

`= 4a - 6a^2 -6 + 9a`

`=  -6a^2 + 13a - 6`