A=1999994 . 1999999 . 1999992 - 1999996 . 1999991 . 1999998>B=23453 . 23458 . 23451 - 23455 . 23450 - 23457

=10198603.91304348 nhé

Chịu ko bt, số j mà dài zữ????

Có thể kb!

#Girl 2k6#

=204835293526529562066306418 nhé

ht

Qua nhà bác google có mấy cái công cụ tính đó

Kb hè

Co minh ne

Bài 1:

a) \(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\)

\(=2^{16}-1\)

b) Sửa đề \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)

\(=\left(3^{32}-1\right)\left(3^{32}+1\right)-3^{64}\)

\(=3^{64}-1-3^{64}\)

\(=-1\)

Bài 2:

Ta có:

\(A=2009.2009\)

\(A=2009\left(2008+1\right)\)

\(A=2009.2008+2009\)

Ta lại có:

\(B=2008.2010\)

\(B=2008\left(2009+1\right)\)

\(B=2008.2009+2008\)

Vì 2008.2009 = 2009.2008

2009 > 2008

=> 2008.2009 + 2009 > 2009.2008 + 2008

=> A > B

1,a,(2-1)(2+1)(2²+1)(2⁴+1)(2⁸+1)

=(2²-1)(2²+1)(2⁴+1)(2⁸+1)

=(2⁴-1) (2⁴+1)(2⁸+1)

=(2⁸ -1)(2⁸+1)=2¹⁶-1

2,

A=2009.2009=2009²

B=2008.2010=(2009-1)(2009+1)=2009²-1

Do2009²>2009²-1\(\Rightarrow A>B\)

b.

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)

Bài 2

\( a)4{\left( {x + 1} \right)^2} + {\left( {2x - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) = 11\\ \Leftrightarrow 4\left( {{x^2} + 2x + 1} \right) + 4{x^2} - 4x + 1 - 8\left( {{x^2} - 1} \right) = 11\\ \Leftrightarrow 4{x^2} + 8x + 4 + 4{x^2} - 4x + 1 - 8{x^2} + 8 = 11\\ \Leftrightarrow 4x + 13 = 11\\ \Leftrightarrow 4x = 11 - 13\\ \Leftrightarrow 4x = - 2\\ \Leftrightarrow x = - \dfrac{1}{2} \)

Bài 2:

\( b)\left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + x\left( {x + 2} \right)\left( {2 - x} \right) = 1\\ \Leftrightarrow {x^3} - 27 + x\left( {2 + x} \right)\left( {2 - x} \right) = 1\\ \Leftrightarrow {x^3} - 27 + x\left( {4 - {x^2}} \right) = 1\\ \Leftrightarrow {x^3} - 27 + 4x - {x^3} = 1\\ \Leftrightarrow 4x = 1 + 27\\ \Leftrightarrow 4x = 28\\ \Leftrightarrow x = 7 \)