• Matéria: Matemática
  • Autor: penaalfonso05pdgztx
  • Perguntado 6 anos atrás

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Anexos:

Respostas

respondido por: Anônimo
1

Explicação passo-a-passo:

a)

\sf (a+1)\cdot(a+1)=a^2+a+a+1

\sf (a+1)\cdot(a+1)=\red{a^2+2a+1}

b)

\sf (2x+4)\cdot(2x-4)=(2x)^2-4^2

\sf (2x+4)\cdot(2x-4)=\red{4x^2-16}

c)

\sf \left(\dfrac{m}{4}-3)\cdot\left(\dfrac{m}{4}+3)=\left(\dfrac{m}{4}\right)^2-3^2

\sf \left(\dfrac{m}{4}-3)\cdot\left(\dfrac{m}{4}\right)=\red{\dfrac{m^2}{16}-9}

d)

\sf (m-2n)\cdot(m+2n)=m^2-(2n)^2

\sf (m-2n)\cdot(m+2n)=\red{m^2-4n^2}

e)

\sf (x^2y+4y)\cdot(x^2y-4y)=(x^2y)^2-(4y)^2

\sf (x^2y+4y)\cdot(x^2y-4y)=\red{x^4y^2-16y^2}

f)

\sf (5x-4)\cdot(5x+4)=(5x)^2-4^2

\sf (5x-4)\cdot(5x+4)=\red{25x^2-16}

g)

\sf (x+y)\cdot(x-y)=\red{x^2-y^2}

h)

\sf \left(xy-\dfrac{z}{2}\right)\cdot\left(xy+\dfrac{z}{2}\right)=(xy)^2-\left(\dfrac{z}{2}\right)^2

\sf \left(xy-\dfrac{z}{2}\right)\cdot\left(xy+\dfrac{z}{2}\right)=\red{x^2y^2-\dfrac{z^2}{4}}

respondido por: Makaveli1996
1

Oie, Td Bom?!

a)

 = (a + 1) \: . \: (a + 1)

 = (a + 1) {}^{2}

 = a {}^{2}  + 2a \: . \: 1 + 1 {}^{2}

 = a {}^{2}  + 2a + 1

b)

 = (2x + 4) \: . \: (2x - 4)

 = (2x) {}^{2}  - 4 {}^{2}

 = 4x {}^{2}  - 16

c)

 = ( \frac{m}{4}  - 3) \: . \: ( \frac{m}{4}  + 3)

 = ( \frac{m}{4} ) {}^{2}  - 3 {}^{2}

 =  \frac{m {}^{2} }{16}  - 9

d)

 = (m - 2n) \: . \: (m + 2n)

 = m {}^{2}  - (2n) {}^{2}

 = m {}^{2}  - 4n {}^{2}

e)

 = (x {}^{2} y + 4y) \: . \: (x {}^{2} y - 4y)

 = (x {}^{2} y) {}^{2}  - (4y) {}^{2}

 = x { }^{4} y {}^{2}  - 16y {}^{2}

f)

 = (5x - 4) \: . \: (5x + 4)

 = (5x) {}^{2}  - 4 {}^{2}

 = 25x {}^{2}  - 16

g)

 = (x + y) \: . \: (x - y)

 = x {}^{2}  - y {}^{2}

h)

 = (xy -  \frac{z}{2} ) \: . \: (xy +  \frac{z}{2} )

 = (xy) {}^{2}  - ( \frac{z}{2}  ){}^{2}

 = x {}^{2} y {}^{2}  -  \frac{z {}^{2} }{4}

Att. Makaveli1996

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