Question

Desenvolver o quadrado da soma e reduzir termos semelhantes?
A) (x+3)² + x² - 7x

B) (x+2)² - (x + 4)² + 4x + 12

Answer 1

$A)~~ (x+3)^2 + x^{2} - 7x \to\\\\ (x+3) (x+3)+ x^{2} - 7x \to\\\\ x^2+3+3x+9+ x^{2} - 7x \to\\\\ x^2+x^2+6x-7x+9\to\\\\ \large\boxed{2x^2-x+9}$




$B) (x+2)^2 - (x + 4)^2 + 4x + 12 \to\\\\ (x+2)(x+2)-(x + 4)(x + 4)+ 4x + 12 \to\\\\(x^2+2x+2x+4)-(x^2+4x+4x+16)+ 4x + 12 \to\\\\ (x^2+4x+4)-(x^2+8x+16)+ 4x + 12 \to\\\\ x^2+4x+4-x^2-8x-16+ 4x + 12 \to\\\\ x^2-x^2+4x-8x+4x+4-16+12\to\\\\ x^2-x^2+8x-8x+16-16\to\\\\ \large\boxed{0}$

Answer 2

a)
= (x + 3)² + x² - 7x 
= x² + 2.3x + 9 + x² - 7x
= x² + 6x + 9 + x² - 7x
= x² + x² + 6x - 7x + 9
= 2x² - x + 9 
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b)
= (x + 2)² - (x + 4)² + 4x + 12 
= x² + 2.2x + 4 - (x² + 2.4x + 16) + 4x + 12
= x² + 4x + 4 - (x² + 8x + 16) + 4x + 12
= x² + 4x + 4 - x² - 8x - 16 + 4x + 12
= x² - x² + 4x - 8x + 4x + 4 - 16 + 12
= 0 + 8x - 8x + 4 + 12 - 16
= 0 + 0 + 16 - 16
= 0