=(2^2-1)*(2^2+1)*(2^4+1)*...*(2^64+1)
=(2^4-1)(2^2+1)*(2^4+1)*...*(2^64+1)
=(2^8-1)(2^8+1)*...*(2^64+1)
=...
=(2^64-1)(2^64+1)
=2^128-1

mk vua lm ra do!!!!!!

tks bạn nhưng mk làm ra lâu rồi sorry nha 

(mặc dù mk vẫn tk cho bạn)

\(=\)\(\left(2^2-1\right)\times\)\(\frac{2^4-1}{2^2-1}\)\(\times\frac{2^8-1}{2^4-1}\)\(\times\)...\(\times\)\(\frac{2^{128}-1}{2^{64}-1}\)

\(=\)\(2^{128}\)\(-1\)

a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)

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

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

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

\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)

\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)

b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)

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

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

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

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

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)

\(=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

\(=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)

\(=\dfrac{3}{2}\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)

\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)

\(C=2-\sqrt{3}+3+\sqrt{3}=5\)

\(A=\dfrac{-4}{9x^2-4}+\dfrac{2x+1}{3x-2}-\dfrac{1}{3x+2}\)

\(=\dfrac{-4+6x^2+4x+3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{6x^2+4x}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{2x}{3x-2}\)

1. -4x( x + 3 )( x - 4 ) - 3x( x² - x + 1 )

= -4x( x² - x - 12 ) - 3x( x² - x + 1 )

= -4x³ + 4x² + 48x - 3x³ + 3x² - 3x

= -7x³ + 7x² + 45x

2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5

<=> 4x² - 20x - ( 4x² - 7x + 3 ) = 5

<=> 4x² - 20x - 4x² + 7x - 3 = 5

<=> -13x - 3 = 5

<=> -13x = 8

<=> x = -8/13

b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4

<=> 6( x² - 7x + 12 ) - 6x² + 12x = 4

<=> 6x² - 42x + 72 - 6x² + 12x = 4

<=> -30x + 72 = 4

<=> -30x = -68

<=> x = 34/15

Bài 1 : 

\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)

\(=-7x^3+7x^2+45x\)

Bài 2 : 

a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)

\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)

\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)

b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)

\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)

\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)