Question

Chứng minh đẳng thức:

1) (xy+z)² -x²y² = z(2xy+z)

2) (x²+y²)² -4x²y² = (x+y)² (x-y)²

Answer 1

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

Biến đổi VT :\(\left(xy+z\right)^2-x^2y^2\)

\(=x^2y^2+2xyz+z^2-x^2y^2\)

\(=2xyz+z^2\)

\(=z\left(2xy+z\right)\) = VP

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

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

Biến đổi VT: \(\left(x^2+y^2\right)^2-4x^2y^2\)

\(=x^4+2x^2y^2+y^4-4x^2y^2\)

\(=x^4-2x^2y^2+y^4\)

Biến đổi VP: \(\left(x+y\right)^2\left(x-y\right)^2\)

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

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

Ta có VT = VP

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

Answer 2

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

\(=x^2y^2+2xyz+z^2-x^2y^2\)

\(=2xyz+z^2\)

\(=z\left(2xy+z\right)=VP\left(đpcm\right)\)

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

\(=x^4+2x^2y^2+y^4-4x^2y^2\)

\(=x^4+y^4-2x^2y^2\)

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

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

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