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6 tháng 8 2019

1a) -3x2(2x3 - 2x + 1/3) = -6x5 + 6x3 - x2

b) (x4 + 2x3 - 2/3).(-3x4) = -3x8 - 6x7 + 2x4

c) (x + 3)(x - 4) = x2 - 4x + 3x - 12 = x2 - x - 12

d)(x - 4)(x2 + 4x + 16) = (x - 4)(x2 + 4x + 42) = x3 - 64

e) 4(x - 1/2)(x + 1/2)(4x2 + 1) =4(x2 - 1/4)(4x2  + 1) = 4(4x4 + x2 - x2 - 1/4) = 4(4x4 - 1/4) = 16x4 - 1

B2. a) (2 - x)(x2 + 2x + 4) + x(x - 3)(x + 4) - x2 + 24 = 0

=> 8 - x3 + x(x2 + 4x - 3x - 12) - x2 + 24 = 0

=> 8 - x3 + x3 + x2 - 12x - x2 + 24 = 0

=> -12x + 32 = 0

=> -12x = -32

=> x = -32 : (-12) = 8/3

b) (x/2 + 3)(5 - 6x) + (12x - 2)(x/4 + 3) = 0

=> 5x/2 - 3x2 + 15 - 18x + 3x2 + 36x - x/2 - 6 = 0

=> 20x + 9 = 0

=> 20x = -9

=> x = -9/20

18 tháng 7 2017

\(A=4.\left(3^2+1\right).\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(=\frac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(=\frac{3^{32}-1}{2}< 3^{32}-1=B\)

Vậy \(A< B\)

a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)

\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)

\(=\dfrac{4x^2-1}{\left(2x-1\right)^3}=\dfrac{2x+1}{\left(2x-1\right)^2}\)

b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)

\(=\dfrac{32}{1+x^{32}}\)

22 tháng 8 2018

Xét :\(D=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\)

             \(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right).\frac{1}{2}\)

                 \(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right).\frac{1}{2}\)

                   \(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+2\right).\frac{1}{2}\)

                     \(=\left(3^8-1\right)\left(3^8+1\right).\frac{1}{2}\)

                       \(=\left(3^{16}-1\right).\frac{1}{2}\)

Vì \(\frac{3^{16}-1}{2}< 3^{16}-1\)

nên D < C

Vậy D < C

8 tháng 7 2019

3. ( 22 + 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1

= ( 22 - 1 ).( 22 + 1 ).( 24 + 1 ).( 28 + 1 )....( 264 + 1 ) + 1

= ( 24 - 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1

= ( 28 + 1 ).....( 264 + 1 )  + 1

= ( 264 - 1 ).( 264 + 1 ) + 1

=  2128 - 1 + 1

= 2128

8 tháng 7 2019

8.( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 32 - 1 ).( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 34 - 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 38 - 1 ).( 38 + 1 )....( 3128 + 1 ) + 1

= ( 316 - 1 )......( 3128 + 1 ) + 1

= ( 3128 - 1 ).( 3128 + 1 ) + 1

=  3256 - 1 + 1

= 3256