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

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

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

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

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

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

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

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

\(A=\left(x+y\right).\left(x+y\right)^2\)

\(A=\left(x+y\right)^3\)

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

Bài giải:

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

a)

\(\begin{array}{l}\left( {2x - 5y} \right)\left( {2x + 5y} \right) + {\left( {2x + 5y} \right)^2}\\ = \left( {2x + 5y} \right)\left( {2x - 5y + 2x + 5y} \right)\\ = \left( {2x + 5y} \right).4x\\ = 2x.4x + 5y.4x\\ = 8{x^2} + 20xy\end{array}\)

b)

\(\begin{array}{l}\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\ = {x^3} + {\left( {2y} \right)^3} + {\left( {2x} \right)^3} - {y^3}\\ = {x^3} + 8{y^3} + 8{x^3} - {y^3}\\ = \left( {{x^3} + 8{x^3}} \right) + \left( {8{y^3} - {y^3}} \right)\\ = 9{x^3} + 7{y^3}\end{array}\)

1)  2xy²+x²y⁴+1=(xy²)²+2xy².1+1²=(xy²+1)²

2)

a)2(x-y)(x+y)+(x+y)²+(x-y)²=(x+y+x-y)²=(2x)²=4x²

b)(x-y+z)²+(z-y)²+2(x-y+z)(y-z)

=(x-y+z)²+(y-z)²+2(x-y+z)(y-z)

=(x-y+z+y-z)²

=x²

a)A=x³+x²y+y²x+y³+2x²y+2xy²

=x³+3x²y+3xy²+y³

A=(x+y)³

b)=3x²+2x+(x²+2x+1)-(4x²-25)=12

3x²+2x+x²+2x+1-4x²+25=12

4x+26=12

= >4x=6/13

= >x=6,5

\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)

\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)

Check lại :v

\(\begin{array}{l}\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) + \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\ = {x^3} - {\left( {2y} \right)^3} + {x^3} + {\left( {2y} \right)^3}\\ = {x^3} - 8{y^3} + {x^3} + 8{y^3}\\ = 2{x^3}\end{array}\)