A=4(3^2+1)(3^4+1)(3^8+1)...(3^64+1)

2A=8(3^2+1)(3^4+1)(3^8+1)...(3^64+1)

2A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)...(3^64+1)

2A=(3^4-1)(3^4+1)(3^8+1)...(3^64+1)

2A=(3^8-1)(3^8+1)....(3^64+1)

2A=(3^16-1)...(3^64+1)

......

2A=(3^64-1)(3^64+1)

2A=3^128-1

A=(3^128-1)/2

=> A>B

\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^8-1\right)\left(3^8+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^{16}-1\right)\left(3^{16}+1\right)...\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)

\(\Leftrightarrow4A=\left(3^{64}-1\right)\left(3^{64}+1\right)\Leftrightarrow4A=3^{128}-1\Leftrightarrow A=\frac{3^{128}-1}{4}\)

Ta có \(\frac{3^{128}-1}{4}< 3^{128}-1\Rightarrow A< B\)

Lâm Huyền:Bạn sai đề rồi B phải là 3¹²⁸-1 chứ !

Sorry,mình tính sai.Bạn thay 4A thành 2A va các số 4 thành số 2 nhé

a/

\(\left(-\frac{1}{16}\right)^{1000}=\left(-\frac{1}{2^4}\right)^{1000}=\left(-\frac{1}{2}\right)^{4000}.\)

Do \(\left(\frac{1}{2}\right)^{4000}>\left(\frac{1}{2}\right)^{5000}\Rightarrow\left(-\frac{1}{2}\right)^{4000}< \left(-\frac{1}{2}\right)^{5000}\Rightarrow\left(-\frac{1}{16}\right)^{1000}< \left(-\frac{1}{2}\right)^{5000}\)

b/

\(3^{400}=\left(3^4\right)^{100}=81^{100}\)

\(4^{300}=\left(4^3\right)^{100}=64^{100}\)

\(\Rightarrow81^{100}>64^{100}\Rightarrow3^{400}>4^{300}\)

khong biet

x=4, y=6,z=8

Trần Thảo Trang

Áp dụng tính chất dãy tỉ số bằng nhau,ta có: 
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x+y+z}{2+3+4}=\frac{18}{9}=2\)

\(\frac{x}{2}=2\Rightarrow x=2.2=4\)

\(\frac{y}{3}=2\Rightarrow y=2.3=6\)

\(\frac{z}{4}=2\Rightarrow z=2.4=8\)

Vậy x = 4, y = 6 và z = 8 

b: \(=x^3-25x+6x-30=\left(x-5\right)\left(x^2+5x+6\right)\)

\(=\left(x-5\right)\left(x+2\right)\left(x+3\right)\)

c: \(=x^3-x-6x-6=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2-x-6\right)\)

\(=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)

d: \(=x^4+4y^4+4x^2y^2-4x^2y^2\)

\(=\left(x^2+2y^2\right)^2-4x^2y^2\)

\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)

e: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)

\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)

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

\(3/4 + 2/1 = 3/4 + 8 / 4 = 11/4\)

\(2/3 - 3/8 = 16/24 - 9 / 24 = 7 /24\)

2/9 . 3/8 = 6 /72

18/1 . 3 / 2 = 54 / 2

a)\(=\dfrac{3}{4}+\dfrac{8}{4}=\dfrac{11}{4}\)

b)\(=\dfrac{16}{24}-\dfrac{9}{24}=\dfrac{7}{24}\)

c)\(=\dfrac{2\times3}{9\times8}=\dfrac{6}{72}=\dfrac{1}{12}\)

d)\(=18\times\dfrac{3}{2}=\dfrac{54}{2}=27\)

11/4

7/24

 6 /72

 54 / 2

>

sp nha

Trả lời :

Ta có :

63¹⁵ < 64¹⁵ = (2⁶)¹⁵ = 2⁹⁰ = (2⁵)¹⁸ = 32¹⁸ < 33¹⁸

=> 63¹⁵ < 33¹⁸

              Bài làm :

Ta có :

63¹⁵ < 64¹⁵ = (2⁶)¹⁵ = 2⁹⁰ = (2⁵)¹⁸ = 32¹⁸ < 33¹⁸

=> 63¹⁵ < 33¹⁸