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d) (8a3 – 27b3) – 2a(4a2 – 9b2)
= (2a – 3b)(4a2 + 6ab + 9b2) – 2a(2a – 3b)(2a + 3b)
= (2a – 3b)(4a2 + 6ab + 9b2 – 4a2 – 6ab) = 9b2(2a – 3b)
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a) Áp dụng HĐT 5 thu được ( 2 a - 3 b ) 3 .
b) Ta có 8 x 3 + 12 x 2 y + 6 xy 2 + y 3 = ( 2 x + y ) 3 .
Áp dụng HĐT 7 với A = 2x + y; B = z
( 2 x + y ) 3 - z 3 = (2x + y - z)(4 x 2 + y 2 + z 2 + 4xy + 2xz + zy).
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a. x3 + 8 = x3 + 23
= (x+2)(x2 - 2x + 4)
b. (x+1)(x2 - x + 1)
= x3 + 13 = x3 + 1
a, Ta có : x^3+8=x^3+2^3=(x+2)(x^2-2x+4)
b, Ta có: (x+1)(x^2-x+1)=x^3+1
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\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
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\(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
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x2 - 2 = \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
\(y^2-13=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)
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\(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
\(125x^3+y^3=\left(5x+y\right)\left(25x^2-5xy+y^2\right)\)
=\(a^3+3^3.b^3=a^3+\left(3+b\right).\left(3^2-3b+b^2\right)\)
`a^3 +27b^3`
`=a^3 + (3b)^3`
`=(a+b) (a^2 - 3ab + b^2)`
Vận dụng HĐT : `A^3 +B^3=(A+B)(A^2 - AB+B^2)`