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\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
\(=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+y^2+2xy\right)\)
\(=\left(x+y\right)\left(x+y\right)^2=\left(x+y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)
\(A=\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(A=\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(A=\left(x+y\right).\left(x+y\right)^2\)
\(A=\left(x+y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
\(\begin{array}{l}\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\ = {x^3} + {\left( {2y} \right)^3} + {\left( {2x} \right)^3} - {y^3}\\ = {x^3} + 8{y^3} + 8{x^3} - {y^3}\\ = \left( {{x^3} + 8{x^3}} \right) + \left( {8{y^3} - {y^3}} \right)\\ = 9{x^3} + 7{y^3}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\begin{array}{l}\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) + \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\ = {x^3} - {\left( {2y} \right)^3} + {x^3} + {\left( {2y} \right)^3}\\ = {x^3} - 8{y^3} + {x^3} + 8{y^3}\\ = 2{x^3}\end{array}\)
\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)
\(\Leftrightarrow A=\left(x+y\right)\left(x^2+2xy+y^2\right)=\left(x+y\right)\left(x+y\right)^2=\left(x+y\right)^3\)
\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
\(\Leftrightarrow A=\left(x^2+y^2\right)\left(x+y\right)+2xy\left(x+y\right)\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)\left(x+y\right)\)
\(\Leftrightarrow A=\left(x+y\right)^2\left(x+y\right)\)
\(\Leftrightarrow A=\left(x+y\right)^3\)