Question

Chứng minh: 

a, (x + y)³ + (x - y)³ = 2x (x² + 3y²)

b, (x + y)³ - (x - y)³ = 2y (3x² + y²)

Giúp tớ với ạ! 

 

Answer 1

a) VT = x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³

          = 2x³ + 6xy²

          = 2x( x² + 3y² ) = VP

=> đpcm

b) VT = x³ + 3x²y + 3xy² + y³ - ( x³ - 3x²y + 3xy² - y³ )

          = x³ + 3x²y + 3xy² + y³ - x³ + 3x²y - 3xy² + y³

          = 3x²y + 2y³

          = 2y( 3x² + y² ) = VP

=> đpcm

Answer 2

a)

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

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

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

b)

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

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

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

Answer 3

a) \(\left(x+y\right)^3+\left(x-y\right)^3\)

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

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

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

b) \(\left(x+y\right)^3-\left(x-y\right)^3\)

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

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

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

Answer 4

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

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

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

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

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

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