Question

Chứng minh các hằng đẳng thức sau:

a) (a - b)^2 = (a + b)^2 - 4ab

b) (x + y)^2 + (x - y)^2 = 2(x^2 + y^2)

Answer 1

a) Ta có:

\(VT=\left(a-b\right)^2\)

\(=a^2-2\cdot a\cdot b+b^2\)

\(=a^2-2ab+b^2\)

\(=a^2-4ab+2ab+b^2\)

\(=\left(a^2+2ab+b^2\right)-4ab\)

\(=\left(a+b\right)^2-4ab=VP\)

⇒ Đpcm

b) Ta có:

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

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

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

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

\(=2x^2+0+2y^2\)

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

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

⇒ Đpcm

Answer 2

a: (a-b)^2

=a^2-2ab+b^2

=a^2+2ab+b^2-4ab

=(a+b)^2-4ab

b: (x+y)^2+(x-y)^2

=x^2+2xy+y^2+x^2-2xy+y^2

=2x^2+2y^2

=2(x^2+y^2)