Question

P = (x + 6)2 – (x + 5)(x – 5) – 3(4x – 1)

F = (x + 3)2 – (x – 3)2 – 2(3x – 8)

con procedimiento xd

Answer 1

Respuesta:

no se bien si algunos de los "2" son exponentes o se multiplican...

En caso de que sean exponentes:

p= (x+6)²-(x+5)(x-5)-3(4x-1)

p= [(x)(x)+2(x)(6)+(6)(6)]-[(x)(x)+(x)(5)+(-5)(x)+(5)(-5)]-12x+3

p=x²+12x+36-(x²-25)-12x+3

p=x²+12x+36-x²+25-12x+3

p=(x²-x²)+(12x-12x)+(36+25+3)

p=64

f=(x+3)²-(x-3)²-2(3x-8)

f=[(x)(x)+2(x)(3)+(3)(3)]-[(x)(x)+2(x)(-3)+(-3)(-3)]-6x+16

f=x²+6x+9-(x²-6x+9)-6x+16

f=x²+6x+9-x²+6x-9-6x+16

f=(x²-x²)+(6x+6x-6x)+(9+-9+16)

f=6x+16