Question

Descompuneti in factori
x²-2=
x²-3=
4x²-5=
2a²-3b²=
1-3²=
9x²-6y²=
12a²-6y²=
25y²-8x²=
ajutati-ma va rog frumos

Answer 1

x²-2=x²-(√2)²=(x-√2)(x+√2)
procedand analog avem

x²-3=(x-√3)(x+√3)

4x²-5=(2x)²-(√5)²= (2x-√5)(2x+√5)

2a²-3b²=(a√2-b√3) (a√2+b√3)

1-3²=1²-3²=(1-3)(1+3)=-2*4=-8; intr-adevar 1-9=-8

9x²-6y²=(3x-y√6)(3x+y√6)


12a²-6y²==6(2a²-y²)=6(a√2-y) (a√2+y)


25y²-8x²=(5y)²-(2√2x)²=(5y-2√2x)(5y+2√2x)