Question

Aplicando as regras dos produtos notáveis, calcule

a) (3x - 2y)²
b) (5x + 8)²
c) (3x² - 2) . (3x³ + 2)
d) (20a5 - 10)²
e) (1/7 + 4x5)²
f) (6x² + 11) (6x² - 11)
g) (10x - 8) (10x + 8)
h) (2x/3 + 10)²
i) (ba² - 3b³)²
j) (9x² - 8x³)²
k) (4x²/3 - 9)
l) (8x + 12)

Answer 1

Bom dia, Luellenlj!

a) $(3x-2y)^2=(3x)^2-2\cdot3x\cdot2y+(2y)^2=9x^2-12xy+4y^2$

b) $(5x+8)^2=(5x)^2+2\cdot5x\cdot8+8^2=25x^2+80x+64$

c) $(3x^2-2)\cdot(3x^2+2)=(3x)^2-2^2=9x^2-4$

d) $(20a^5-10)^2=(20a^5)^{2}-2\cdot20a^5\dot10+10^2=400a^{10}+400a^5+100$

e) $\left(\dfrac{1}{7}+4x^5\right)^2=\left(\dfrac{1}{7}\right)^2+2\cdot\dfrac{1}{7}\cdot4x^5+(4x^5)^2=\dfrac{1}{49}+\dfrac{8x^5}{7}+16x^{10}$

f) $(6x^2+11)(6x^2-11)=(6x)^2-11^2=36x^4-121$

g) $(10x-8)(10x+8)=(10x)^2-8^2=100x^2-64$

h) $\left(\dfrac{2x}{3}+10\right)^2=\left(\dfrac{2x}{3}\right)^2+2\cdot\dfrac{2x}{3}\cdot10+10^2=\dfrac{4x^2}{9}+\dfrac{40x}{3}+100$

i) $(ba^2-3b^3)^2=(ba^2)^2-2\cdot ba^2\cdot3b^3+(3b^3)^2=b^2a^4-6a^2b^4+9b^6$

j) $(9x^2-8x^3)^2=(9x^2)^2-2\cdot9x^2\cdot8x^3+(8x^3)^2=81x^4-144x^5+64x^6$

k) $\left(\dfrac{4x^2}{3}-9\right)^2=\left(\dfrac{4x^2}{3}\right)^2-2\cdot\dfrac{4x^2}{3}\cdot9+9^2=\dfrac{16x^4}{9}-24x^2+81$

l) $(8x+12)^2=(8x)^2+2\cdot8x\cdot12+12^2=64x^2+192x+144$