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

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

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

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

=> 8F = \(\left(3^8-1\right)\left(3^8+1\right)-3^{16}=3^{16}-1-3^{16}=-1\)

=> F = -1/8

b) Ta có G = \(\left(2^3+1\right)\left(2^6+1\right)\left(2^{12}+1\right)-\frac{2^{24}}{7}\)

=> 7G = 7(2³ + 1)(2⁶ + 1)(2¹² + 1) - 2²⁴

=> 7G = (2³ - 1)(2³ + 1)(2⁶ + 1)(2¹² + 1) - 2²⁴

=> 7G = (2⁶ - 1)(2⁶ + 1)(2¹² + 1) - 2²⁴

=> 7G = (2¹² - 1)(2¹² + 1) - 2²⁴

=> 7G = 2²⁴ - 1 - 2²⁴

=> 7G = -1

=>  G = -1/7

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

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

<=> \(8F=\left(3^4-1\right)\left(3^4+1\right)\left(3^8-1\right)-3^{16}\)

<=> \(8F=\left(3^8+1\right)\left(3^8-1\right)-3^{16}\)

<=> \(8F=\left(3^{16}-1\right)-3^{16}=-1\)

<=> F = -1/8

Câu G làm tương tự

nhân hết ra rồi tính

a: \(2^2\cdot5-\dfrac{\left(1^{10}+8\right)}{3^2}\)

\(=4\cdot5-\dfrac{1+8}{3}\)

=20-3

=17

b: \(5^8:5^6+4\left(3^2-1\right)\)

\(=5^2+4\left(9-1\right)\)

=25+4*8

=25+32

=57

c: \(400-\left\{36-20:\left[3^3-\left(8-3\right)\right]\right\}\)

\(=400-36+20:\left[27-5\right]\)

\(=364+\dfrac{20}{22}\)

\(=364+\dfrac{10}{11}=\dfrac{4014}{11}\)

A) 2².5 - (1¹⁰ + 8) : 3²

= 4.5 - (1 + 8) : 9

= 20 - 9 : 9

= 20 - 1

= 19

B) 5⁸ : 5⁶ + 4.(3² - 1)

= 5² + 4.(9 - 1)

= 25 + 4.8

= 25 + 32

= 57

C) 400 - {36 - 20 : [3³ - (8 - 3)]}

= 400 - [36 - 20 : (27 - 5)]

= 400 -(36 - 20 : 22)

= 400 - (36 - 10/11)

= 400 - 386/11

= 4014/11

\(A=16^8-1=2^{32}-1\)

\(\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right)=\left(2.2^2.2^4.2^8.2^{16}\right).\left(1+1+1+1+1\right)=\left(2^{31}\right).5\)

a) \(\left(8^{2019}-8^{2018}\right):\left(8^{2016}.8^2\right)\)

\(=8^{2018}\left(8-1\right):8^{2016+2}\)

\(=8^{2018}.7:8^{2018}=7\)

b) Em tham khảo link : Câu hỏi của ✽❤Girl cute❤✽ - Toán lớp 6 - Học toán với OnlineMath

em cảm ơn chị ạ!

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

B = \(\left(-1\dfrac{1}{6}\right)\) : \(\left(\dfrac{-10}{3}+\dfrac{9}{4}\right)\) - \(\left(-\dfrac{3}{8}\right)\) : \(\left(8-\dfrac{51}{8}\right)\)

B = \(\dfrac{-7}{6}\) : \(\dfrac{-13}{12}\) - \(\left(-\dfrac{3}{8}\right)\) : \(\dfrac{13}{8}\)

B = \(\dfrac{14}{13}\) - \(\dfrac{-3}{13}\)

B = \(\dfrac{17}{13}\)

Sửa đề

\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{1024}+1\right)+1\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{1024}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{1024}+1\right)+1\)

\(..................................................\)

\(=\left(2^{1024}-1\right)\left(2^{1024}+1\right)+1\)

\(=2^{2048}-1+1=2^{2048}\)

Phải là (2+1)(2²+1)(2⁴+1)...(2³²+1)- 2^64

(2+1)(2²+1)(2⁴+1)...(2³²+1)

=(2-1)(2+1)(2²+1)(2⁴+1)...(2³²+1)

=(2²-1)(2²+1)(2⁴+1)...(2³²+1)

=(2⁴-1)(2⁴+1)...(2³²+1)=…=2^64-1

Vậy C=-1

Ta có: \(\left(-2\right)^2\cdot3-\left(1^{10}+8\right):\left(-3\right)^2\)

\(=4\cdot3-\dfrac{9}{9}\)

=12-1=11

\(\left(-2\right)^2.3-\left(1^{10}+8\right):\left(-3\right)^2\)

\(=4.3-9:9\)

\(=12-1=11\)

(-2)².3 - (1¹⁰ + 8) : ( -3)²

= 4.3 - ( 1 + 8) : 9

= 12 - 9 : 9

= 12 -1 

= 11