Exercícios resolvidos de produtos notáveis.

Exercícios de produtos notáveis.

  1. Usando a regra do quadrado da soma de dois números, obtenha os trinômios quadrados perfeitos que resultam das expressões a seguir. $${(uv + z)}^2 $$ $$ {(5m + r)}^2 $$ $$ {(7 + 2p)}^2$$ $${(a + 6b)}^2$$ $${(10x^{2 }+ y^{2})}^2$$ $${(mp^{3} + nr^{2})}^2$$

 

$${(uv + z)}^2$$ $${(uv)^2 + 2\cdot{uv}\cdot{z} + {z}^2} $$ $$ {u^2 v^2} + 2uvz + z^2 $$

 

$${(5m + r)}^2$$ $$ (5m)^{2} + 2\cdot{5m}\cdot r + r^{2} $$ $$ 25m^2 +10mr + r^2 $$

 

$$ {(7 + 2p)}^2$$ $$ 7^2 + 2\cdot 7\cdot {2p} + {^(2p)}^2 $$ $$ 49 + 28p +4p^2 $$

 

$${(a + 6b)}^2$$ $${ a^2 + 2\cdot a\cdot{6b} + {(6b)}^2 }$$ $$ a^2 + 12ab + 36b^2 $$

 

$${(10x^{2 }+ y^{2})}^2$$ $$ {(10x^{2}}^2 + 2\cdot{(10x^{2}}\cdot{y^{2}} + {y^{2}}^2 $$ $$ 100x^4 + 20x^{2}y^{2} + y^{4} $$

 

$${(mp^{3} + nr^{2})}^2$$ $${(mp^{3})}^2 +2\cdot{(mp^3)}\cdot (nr^{2}) + {(nr^{2})}^2 $$ $$ m^{2}p^{6} + 2mnp^{3}r^{2} + n^{2}r^{4} $$

 

2. Faça o mesmo usando a regra do quadrado da diferença entre dois números, com as expressões abaixo. $${(5a – 2b)}^2$$ $$ {(a^{2}i – b^{3}j)}^2$$ $$ {(2vx – 3uy)}^2$$ $$ {(4 q^{3} – 6p^{2})}^{2} $$ $${(12 – 3 a^{3})}^2$$ $$ {(15 – 3x)}^2$$ $$ {(7x – 8y)}^2 $$

 

$${(5a – 2b)}^2$$ $$ {(5a)}^2 – 2\cdot {5a}\cdot{2b} +{(ab)}^2 $$ $$ 25a^2 – 20ab + 4b^2 $$

 

$$ {(a^{2}i – b^{3}j)}^2$$ $$ [(a^2)i]^2 – 2\cdot{a^2}i\cdot{b^3} + {(b^3)}^2 $$ $$ a^{4}i^{2} – 2a^{2}b^{3}i + b^6 $$

 

$$ {(2vx – 3uy)}^2$$ $${(2vx)^2 – 2\cdot {(2vx)}\cdot{(3uy)} + {(3uy)}^2} $$ $$ 4v^{2}x^{2} – 12uvxy + 9u^{2}y^{2} $$

 

$$ {(4 q^{3} – 6p^{2})}^{2} $$ $$ (4q^{3})^2 – 2\cdot (4q^{3})\cdot(6p^{2}) +{(6p^{2})}^2 $$ $$ 16q^6 – 48q^{3}p^{2} + 36p^{4} $$

 

$${(12 – 3 a^{3})}^2$$ $${(12)^2 – 2\cdot{12}\cdot{3a^{3}} + {(3a^{3})}^2} $$ $$ 144 – 72a^{3} + 9a^6 $$

 

$$ {(15 – 3x)}^2$$ $$ {(15)}^2 – 2\cdot {15}\cdot{(3x)} + {(3x)}^2 $$ $$225 – 90x + 9x^2 $$

 

$$ {(7x – 8y)}^2 $$ $${(7x)}^2 – 2\cdot{(7x)}\cdot 8y + {(8y)}^2 $$ $$ 49x^2 – 112xy + 64y^2$$

 

3. Usando a regra do produto da soma de dois números pela sua diferença, obtenha os binômios resultantes das multiplicações abaixo. $${(7 + 2x)}{(7 – 2x)}$$ $${(5 – 3y)}{(5 + 3y)}$$ $$ {(ab^{2} + b)}{(ab^{2} – b)}$$ $${(xy + xz)}{(xy – xz)}$$ $$ {(4m – 3n)}{(4m + 3n)}$$ $$ {(7x^{3} + 2y^{2})}{(7x^{3} – 2y^{2})}$$

 

$${(7 + 2x)}{(7 – 2x)}$$ $$ {7^2 – {(2x)}^2} $$ $$ 49 – 4x^2 $$

 

$${(5 – 3y)}{(5 + 3y)}$$ $$ 5^2 – {(3y)}^2 $$ $$ 25 – 9y^2 $$

 

$$ {(ab^{2} + b)}{(ab^{2} – b)}$$ $$ {(ab^{2}}^{2} – b^2 $$ $$ a^{2}b^{4} – b{2} $$

 

$${(xy + xz)}{(xy – xz)}$$ $${(xy)}^{2} – {(xz)}^{2} $$ $$ x^{2}y^{2} – x^{2}z^{2} $$

 

$$ {(4m – 3n)}{(4m + 3n)}$$ $$ {(4m)}^{2} – {(3n)}^{2} $$ $$ 16m^2 – 9n^2  $$

 

$$ {(7x^{3} + 2y^{2})}{(7x^{3} – 2y^{2})}$$ $${(7x^{3})}^{2} – {(2y^{2})}^2 $$ $$ 49x^{6} – 4y^{4} $$

 

4. Use agora a regra do cubo da soma de dois números para obter os polinômios de quatro termos resultantes das expressões abaixo.  $${(7 +2j)}^3$$ $$ {(x + 3yz)}^3$$ $$ {(4l + 5m)}^3$$ $${(ma + nb)}^3 $$ $${(11 + 4r)}^3 $$

 

$${(7 +2j)}^3$$ $$ 7^3 + 3\cdot {7^{2}}\cdot{2j} + 3\cdot {7}\cdot {(2j)}^{2} + {(2j)}^3 $$ $$343 + 294j + 84j^2  + 8j^3 $$

 

$$ {(x + 3yz)}^3$$ $$ x^3 + 3\cdot x^{2}\cdot {(3yz)} + 3\cdot x\cdot {(3yz)}^2 + {(3yz)}^3 $$ $$ x^3 + 9x^{2}yz + 27xy^{2}z^{2} + 27y^{3}z^{3} $$

 

$$ {(4l + 5m)}^3$$ $${(4l)}^3 +3\cdot{4l}^{2}\cdot{(5m)} +3\cdot{(4l)}\cdot{(5m)}^2 + {(5m)}^3 $$ $$ 64l^3 + 240l^{2}m + 125m^{3} $$

 

$${(ma + nb)}^3 $$ $${(ma)^3} + 3\cdot {(ma)}^{2}\cdot {(nb)} +3\cdot {(ma)}\cdot {(nb)}^{2} + {(nb)}^{3} $$ $$ m^{3}a^{3} + 3m^{2}na^{2}b + 3mn^{2}ab^{2} + n^{3}b^{3} $$

 

$${(11 + 4r)}^3 $$ $$ 11^3 + 3\cdot 11^{2}\cdot{(4r)} + 3\cdot 11\cdot{(4r)}^{2} + {(4r)}^{3} $$ $$ 1331 + 1452 r + 528 r^2 + 64 r^3 $$

 

5. Vamos fazer o mesmo com a regra do cubo da diferença. $${(4m – 2)}^3$$ $${(3x – 5y)}^3$$ $${(9 – 5a)}^3$$ $${(5 – 4x)}^3 $$ $${(10 – 5c)}^3 $$ $${(3ab – x)}^3$$ $${(pq^{2} – rq)}^3$$

 

$${(4m – 2)}^3$$ $$ {(4m)}^{3} – 3\cdot {(4m)^{2}}\cdot 2 + 3\cdot{(4m)}\cdot 2^{2} – 2^{3} $$ $$ 64m^{3} – 96m^{2} + 48m – 8 $$

 

$${(3x – 5y)}^3$$ $$ (3x)^{3} – 3\cdot (3x)^{2}\cdot {(5y)} + 3\cdot{(3x)}\cdot (5y)^{2} – {(5y)}^{3} $$ $$ 27x^3 – 135x^{2}y + 225xy^{2} – 125y^{3} $$

 

$${(9 – 5a)}^3$$ $$ {(9)^{3}} – 3\cdot (9)^2\cdot (5a) + 3\cdot 9 \cdot(5a)^{2} -(5a)^{3} $$ $$729 – 1215 a + 675 a^2 – 125 a^3 $$

 

$${(5 – 4x)}^3 $$ $$ 5^{3} – 3\cdot 5^{2}\cdot (4x) + 3\cdot 5\cdot (4x)^{2} – {(4x)}^{3} $$ $$125 – 300x + 120 x^{2} – 64x^{3} $$

 

$${(10 – 5c)}^3 $$ $$ 10^{3} – 3\cdot(5)^{2}\cdot (5c) + 3\cdot{10}\cdot {(5c)}^{2} – {(5c)}^{3} $$ $$ 1000 – 375 c + 750 c^{2} – 125c^{3} $$

 

$${(3ab – x)}^3$$ $${(3ab)}^{3} – 3\cdot {(3ab)}^{2}\cdot x + 3\cdot{(3ab)}\cdot x^{2} – x^{3} $$ $$ 27a^{3}b^{3} – 27a^{2}b^{2}x + 9 abx^{2} – x^{3} $$

 

$${(pq^{2} – rq)}^3$$ $$ {(pq^{2})}^{3} – 3\cdot {(pq^{2})}^{2}\cdot rq + 3\cdot {(pq^{2})}\cdot {(rq)}^{2} – {(rq)}^{3} $$ $$ p^{3}q^{6} – 3p^{2}q^{5}r + 3pq^{4}r^{2} – q^{3}r^{3} $$

 

6. Chegou o momento de usar as regras mais avançadas. Multiplique os quadrados das somas pelas diferenças dos mesmos números, usando a regra vista no post anterior. $${(ax + by)}^{2}\cdot {(ax – by)} $$ $$ {(5 + 3x)}^{2}\cdot{(5 – 3x)} $$ $$ {(4n + m^{2})}^{2}\cdot{(4n – m)} $$ $${(5a + 3b)}^{2}\cdot{(5a – 3b)} $$ $${(7x + 2y)}^{2}\cdot{7x -2y)} $$ $${(10 + 3v)}^{2}\cdot{(10 – 3v)}$$ $${(px + qy)}^{2}\cdot{(px – qy)} $$

$${(ax + by)}^{2}\cdot {(ax – by)} $$ $${(ax)}^3 + {(ax)}^{2}\cdot{(by)} – ax\cdot {(by)}^{2} – {(by)}^{3} $$ $$ a^{3}x^{3} + a^{2}bx^{2}y – ab^{2}xy^{2} – {(by)}^{3} $$ $$a^{3}x^{3} + a^2bx^2y – ab^2xy^2 – b^3y^3 $$

 

 

$$ {(5 + 3x)}^{2}\cdot{(5 – 3x)} $$ $$ 5^{3} + 5^2\cdot{3x} – 5\cdot {(3x)}^{2} – {(3x)}^{3} $$ $$ 125 + 75x – 45x^2 – 27x^3 $$

 

$$ {(4n + m^{2})}^{2}\cdot{(4n – m)} $$ $$ {(4n)}^{3} + {(4n)}^{2}\cdot{(m^2)} – 4n\cdot {(m^2)}^2 – {(m^2)}^{3} $$ $$ 64n^3 + 16n^2m^2 – 4nm^4 – m^6 $$

 

$${(5a + 3b)}^{2}\cdot{(5a – 3b)} $$ $${(5a)}^{3} +{(5a)}^{2}\cdot {(3b)} – 5a\cdot{(3b)^{2}} – {(3b)}^3 $$ $$125a^3 + 75a^2b – 45ab^2 – 27b^3 $$

 

$${(7x + 2y)}^{2}\cdot{7x -2y)} $$ $$ {(7x)}^{3} + {(7x)}^{2}\cdot {(2y)} – 7x\cdot{(2y)^2} – {(2y)^3} $$ $$ 343x^3 + 98x^2y – 14xy^2 – 8y^3 $$

 

$${(10 + 3v)}^{2}\cdot{(10 – 3v)}$$ $$(10)^3 + (10)^2\cdot {(3v)} – 10\cdot{(3v)}^2 – {(3v)}^3 $$ $$ 1000 + 300v – 90v^2 – 9v^3 $$

 

$${(px + qy)}^{2}\cdot{(px – qy)} $$ $$ {(px)}^3 + {(px)}^2\cdot {(qy)} – px\cdot{(qy)}^2 – {(qy)}^3 $$ $$ p^3x^3 + p^2qx^2y – pq^2xy^2 – q^3y^3 $$

 

7. Agora vamos multiplicar o quadrado das diferenças, pelas somas dos dois números, conforme a regra vista. $${(3x – 2y)}^{2}\cdot{(3x + 2y)} $$ $${(5a – bx)}^{2}\cdot{(5a + bx)}$$ $${(1 – 5x)}^{2}\cdot{(1 + 5x)}$$ $$ {(6t – 4s)}^{2}\cdot{(6t+ 4s)}$$ $${(8l – z)}^{2}\cdot{(8l +z)} $$ $${(4n – 5m)}^{2}\cdot{(4n +5m)}$$ $${(r – pq)}^{2}\cdot{(r + pq)} $$

 

$${(3x – 2y)}^{2}\cdot{(3x + 2y)} $$ $$ {(3x)}^3 – {(3x)}^2\cdot{(2y)} – 3x\cdot {(2y)^2} + {(2y)}^3 $$ $$ 27x^3 – 18x^2y – 12xy^2 + 8y^3 $$

 

$${(5a – bx)}^{2}\cdot{(5a + bx)}$$ $${(5a)}^3 – {(5a)^2}\cdot{(bx)} – 5a\cdot{(bx)}^2 + {(bx)}^3 $$ $$ 125 a^3 – 25abx – 5ab^2x^2 + b^3x^3 $$

 

$${(1 – 5x)}^{2}\cdot{(1 + 5x)}$$ $$ 1^3 – 1^2\cdot 5x – 1\cdot {(5x)^2} + {(5x)}^3 $$ $$ 1 – 5x – 25x^2 + 125x^3 $$

 

$$ {(6t – 4s)}^{2}\cdot{(6t+ 4s)}$$ $${(6t)}^3 – {(6t)}^2\cdot {(4s)} – 6t\cdot {(4s)}^2 + {(4s)}^3 $$ $$ 216t^3 – 144t^2s – 96ts^2 + 64s^3 $$

 

$${(8l – z)}^{2}\cdot{(8l +z)} $$ $${(8l)}^3 – {(8l)^2}\cdot (z) – 8l\cdot z^2 + z^3 $$ $$ 512l^3 – 64l^2z – 8lz^2 + z^3$$

 

$${(4n – 5m)}^{2}\cdot{(4n +5m)}$$ $$ {(4n)}^3 – {(4n)^2}\cdot{(5m} -4n\cdot {((5m)}^2 + {(5m)}^3 $$ $$ 64n^3 – 80mn^2 – 100m^2n + 125m^3 $$

 

$${(r – pq)}^{2}\cdot{(r + pq)} $$ $$ r^3 – r^2\cdot {(pq)} – r\cdot {(pq)}^2 + {(pq)}^3 $$  $$ r^3 – pqr^2 – p^2q^2r + p^3q^3 $$

Curitiba, l8 de abril de 2016

Décio Adams

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