Question

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BINOMIOS

x2 – 4

b2 – 1

x2 - 9/25

36x2 - a6b4

9x2 – 4y4

x4 – 16 9

x4 − 4x2

36x6 − 49

x²- 25

x²- y²

9x²- 4y²

9a2 - 16b2

(7x + 3)2 - (5x - 4)2

125a3 - b3c3

27x3 + 8y3

125a3 + 8b3

x5 + y5

x3 – 8

b4 – 81

x7 + 1

x6 - 1/64

x7 + 128a7

x7 - y7

x3 + 1

x7 - y7

a4 - b4

(x – 1)3 – (1 – x)3

Answer 1

Respuesta:

Explicación:

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

$b^{2} -1= (b+1)(b-1)$

$x^{2} -\frac{9}{25} = (x+\frac{3}{5})(x-\frac{3}{5})$

$36^{2} - a^{6}b^{4}= (6+a^{3} b^{2}) (6-a^{3}b^{2})$

$9x^{2} -4y^{4} = (3x+2y)(3x-2y)$

$x^{4} -169=(x^{2}+13)(x^{2} -13)$

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

$36x^{6}-49=(6x^{3}+7)(6x^{3}-7)$

$x^{2} -25= (x+5)(x-25)$

$x^{2} -y^{2} =(x+y)(x-y)$

$9x^{2} -4y^{2}=(3x-2y)(3x+2y)$

$9a^{2} -16b^{2} =(3a+4b)(3a-4b)$

$(7x+3)^{2} -(5x-4)^{2}= (7x+3+5x-4)(7x+3-5x+4)\\\\=(12x-1)(2x+7)$

$125a^{3} -b^{3} c^{3} =(5a)^{3}-(bc)^{3}\\  =(5a-bc)(25a^{2}+5abc+(bc)^{2})$

$27x^{3}+8y^{3}= (3x)^{3} +(2y)^{3}\\ (3x+2y)(9x^{2} -6xy+4y^{2} )$

$125a^{3}+8b^{3}=(5a)^{3}+(2b)^{3}\\    (5a+2b)(25a^{2}-10ab+4b^{2} )$

$x^{5}+y^{5}  = (x+y)(x^{4}-x^{3}y+x^{2} y^{2}-xy^{3}+y^{4})$

$x^{3} -8=x^{3}-2^{3}\\=(x-2)(x^{2} +2x+4)$

$b^{4}-81= (b^{2}+9)(b^{2} -9)=(b^{2}+9)(b+3)(b-3)$

$x^{7}+1= (x+1)(x^{6}-x^{5}(1)+x^{4}(1)^{2}-x^{3}(1)^{3} +x^{2}(1)^{4}-x^{1} (1)^{5}+ 1^{6} )$

$x^{6} -\frac{1}{64}=(x^{3} -\frac{1}{8})(x^{3}+\frac{1}8})\\    =(x-\frac{1}{2})(x^{2} +\frac{1}{2}x+\frac{1}{4})(x+\frac{1}{2})(x^{2} -\frac{1}{2}x+\frac{1}{4}   )$

$x^{7} -(2a)^{7} =(x-2a)(x^{6}+x^{5}2a+x^{4}(2a)^{2}+x^{3}(2a)^{3}+x^{2}(2a)^{4}+x(2a)^{5}+(2a)^{6}   )$

$x^{7}-y^{7}=(x-y)(x^{6}+x^{5}y+x^{4}y^{2}+x^{3}y^{3}+x^{2} y^{4}+xy^{5}+y^{6})$

$x^{3} +1= (x+1)(x^{2} -x+1)$

$x^{7}-y^{7}=---$

$a^{4}-b^{4}= (a^{2}-b^{2})(a^{2}+b^{2})=(a+b)(a-b)(a^{2}+b^{2})$

$(x-1)^{3}-(1-x)^{3}  =x^{3} -3x^{2} +3x+1-1+3x-3x^{2} -x^{3} \\=6x-6x^{2} =6x(1-x)$