Question

Preciso para agora. dou 50 pontos pra quem responder certo



produto notável de:


a). (0,2x+4y²)
b). (3x+2b)•(3x-2b)
c). (⁸/⁷ x+5y³z)²
d). (2x+⅕ y)
e). (2y-5)³
f). (a²b²+b)²
g). (xz-3)²
h). (a-3x)³
i). (⅚+y)²
j). (a²b-¼)²
k). (⁷/³ x+5)•(⁷/³ x-5)
l). (x²z+¼w)³
​

Answer 1

Explicação passo-a-passo:

a)

$\sf (0,2x+4y^2)^2=(0,2x)^2+2\cdot0,2x\cdot4y^2+(4y^2)^2$

$\sf (0,2x+4y^2)^2=\red{0,04x^2+1,6xy^2+16y^4}$

b)

$\sf (3x+2b)\cdot(3x-2b)=(3x)^2-(2b)^2$

$\sf (3x+2b)\cdot(3x-2b)=\red{9x^2-4b^2}$

c)

$\sf \left(\dfrac{8x}{7}+5y^3z\right)^2=\left(\dfrac{8x}{7}\right)^2+2\cdot\dfrac{8x}{7}\cdot5y^3z+(5y^3z)^2$

$\sf \left(\dfrac{8x}{7}+5y^3z\right)^2=\red{\dfrac{64x^2}{49}+\dfrac{80xy^3z}{7}+25y^6z^2}$

d)

$\sf \left(2x+\dfrac{y}{5}\right)^2=(2x)^2+2\cdot2x\cdot\dfrac{y}{5}+\left(\dfrac{y}{5}\right)^2$

$\sf \left(2x+\dfrac{y}{5}\right)^2=\red{4x^2+\dfrac{4xy}{5}+\dfrac{y^2}{25}}$

e)

$\sf (2y-5)^3=(2y)^3-3\cdot(2y)^2\cdot5+3\cdot2y\cdot5^2-5^3$

$\sf (2y-5)^3=8y^3-3\cdot4y^2\cdot5+3\cdot2y\cdot25-125$

$\sf (2y-5)^3=\red{8y^3-60y^2+150y-125}$

f)

$\sf (a^2b^2+b)^2=(a^2b^2)^2+2\cdot a^2b^2\cdot b+b^2$

$\sf (a^2b^2+b)^2=\red{a^4b^4+2a^2b^3+b^2}$

g)

$\sf (xz-3)^2=(xz)^2-2\cdot xz\cdot3+3^2$

$\sf (xz-3)^2=\red{x^2z^2-6xz+9}$

h)

$\sf (a-3x)^3=a^3-3\cdot a^2\cdot3x+3\cdot a\cdot(3x)^2-(3x)^3$

$\sf (a-3x)^3=a^3-9a^2x+3\cdot a\cdot9x^2-27x^3$

$\sf (a-3x)^3=\red{a^3-9a^2x+27ax^2-27x^3}$

i)

$\sf \left(\dfrac{5}{6}+y\right)^2=\left(\dfrac{5}{6}\right)^2+2\cdot\dfrac{5}{6}\cdot y+y^2$

$\sf \left(\dfrac{5}{6}+y\right)^2=\red{\dfrac{25}{36}+\dfrac{5y}{3}+y^2}$

j)

$\sf \left(a^2b-\dfrac{1}{4}\right)^2=(a^2b)^2-2\cdot a^2b\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2$

$\sf \left(a^2b-\dfrac{1}{4}\right)^2=\red{a^4b^2-\dfrac{a^2b}{2}+\dfrac{1}{16}}$

k)

$\sf \left(\dfrac{7x}{3}+5\right)\cdot\left(\dfrac{7x}{3}-5\right)=\left(\dfrac{7x}{3}\right)^2-5^2$

$\sf \left(\dfrac{7x}{3}+5\right)\cdot\left(\dfrac{7x}{3}-5\right)=\red{\dfrac{49x^2}{9}-25}$

l)

$\sf \left(x^2z+\dfrac{w}{4}\right)^2=(x^2z)^2+2\cdot x^2z\cdot\dfrac{w}{4}+\left(\dfrac{w}{4}\right)^2$

$\sf \left(x^2z+\dfrac{w}{4}\right)^2=\red{x^4z^2+\dfrac{x^2wz}{2}+\dfrac{w^2}{16}}$

Answer 2

Oie, Td Bom?!

a)

$= (0,2x + 4y {}^{2} ) {}^{2}$

$= ( \frac{1}{5} x + 4y {}^{2} ) {}^{2}$

$= ( \frac{1}{5} x) {}^{2} + 2 \: . \: \frac{1}{5} x \: . \: 4y {}^{2} + (4y {}^{2} ) {}^{2}$

$= \frac{1}{25} x {}^{2} + \frac{8}{5} xy {}^{2} + 16y {}^{4}$

b)

$= (3x + 2b) \: . \: (3x - 2b)$

$= (3x) {}^{2} - (2b) {}^{2}$

$= 9x {}^{2} - 4b {}^{2}$

c)

$= ( \frac{8}{7} x + 5y {}^{3} z) {}^{2}$

$= ( \frac{8}{7} x) {}^{2} + 2 \: . \: \frac{8}{7} x \: . \: 5y {}^{3} z + (5y {}^{3} z) {}^{2}$

$= \frac{64}{49} x {}^{2} + \frac{80}{7} xy {}^{3} z + 25y {}^{6} z {}^{2}$

d)

$= (2x + \frac{y}{5} ) {}^{2}$

$= (2x) {}^{2} + 2 \: . \: 2x \: . \: \frac{y}{5} + ( \frac{y}{5} ) {}^{2}$

$= 4x {}^{2} + \frac{4xy}{5} + \frac{y {}^{2} }{25}$

e)

$= (2y - 5) {}^{3}$

$= (2y) {}^{3} - 3 \: . \: (2y) {}^{2} \: . \: 5 + 3 \: . \: 2y \: . \: 5 {}^{2} - 5 {}^{3}$

$= 8y {}^{3} - 3 \: . \: 4y {}^{2} \: . \: 5 + 3 \: . \: 2y \: . \: 25 - 125$

$= 8y {}^{3} - 60y {}^{2} + 150y - 125$

f)

$= (a {}^{2} b {}^{2} + b) {}^{2}$

$= (a {}^{2} b {}^{2} ) {}^{2} + 2a {}^{2} b {}^{2} \: . \: b + b {}^{2}$

$= a {}^{4} b {}^{4} + 2a {}^{2} b {}^{3} + b {}^{2}$

g)

$= (xz - 3) {}^{2}$

$= (xz) {}^{2} - 2xz \: . \: 3 + 3 {}^{2}$

$= x {}^{2} z {}^{2} - 6xz + 9$

h)

$= (a - 3x) {}^{3}$

$= a {}^{3} - 3a {}^{2} \: . \: 3x + 3a \: . \: (3x) {}^{2} - (3x) {}^{3}$

$= a {}^{3} - 9a {}^{2} x + 3a \: . \: 9x {}^{2} - 27x {}^{3}$

$= a {}^{3} - 9a {}^{2} x + 27ax {}^{2} - 27x {}^{3}$

i)

$= ( \frac{5}{6} + y ){}^{2}$

$= ( \frac{5}{6}) {}^{2} + 2 \: . \: \frac{5}{6} y + y {}^{2}$

$= \frac{25}{36} + \frac{5}{3} y + y {}^{2}$

j)

$= (a {}^{2} b - \frac{1}{4} ) {}^{2}$

$= (a {}^{2} b) {}^{2} - 2a {}^{2} b \: . \: \frac{1}{4} + ( \frac{1}{4} ) {}^{2}$

$= a{}^{4} b {}^{2} - \frac{1}{2} a {}^{2} b + \frac{1}{16}$

k)

$= ( \frac{7}{3} x + 5) \: . \: ( \frac{7}{3} x - 5)$

$= ( \frac{7}{3} x) {}^{2} - 5 {}^{2}$

$= \frac{49}{9} x {}^{2} - 25$

l)

$= (x {}^{2} z + \frac{w}{4} ) {}^{2}$

$= (x {}^{2} z) {}^{2} + 2x {}^{2} z \: . \: \frac{w}{4} + ( \frac{w}{4} ) {}^{2}$

$= x{}^{4} z {}^{2} + \frac{2wx {}^{2}z }{4} + \frac{w {}^{2} }{16}$

$= x {}^{4} z {}^{2} + \frac{wx {}^{2}z }{2} + \frac{w {}^{2} }{16}$

Att. Makaveli1996