I am a loser: Bạn chép đề sai nha, mình sửa luôn.

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

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

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

\(A=\left(2^8-1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)

\(A=\left(2^{16}-1\right)\cdot\left(2^{16}+1\right)\)

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

Vậy...

Mình sửa đề bài nha:

\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)

\(=\frac{1}{2}\left(5^{32}-1\right)\)

\(=\frac{5^{32}-1}{2}\)

Chúc bạn học tốt!

bạn tính ra \(\frac{1}{2}\)kiểu gì đấy chỉ mình với

helppppppppppppppppppp

\(4A=4\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5^2-1\right)\left(5^2+1\right).....\left(5^{2048}+1\right)\left(hdt\left(a-b\right)\left(a+b\right)=a^2-b^2\right)=\left(5^4-1\right)\left(5^4+1\right)......\left(5^{2048}+1\right)=\left(5^8-1\right).....\left(5^{2048}+1\right)=.....=\left(5^{1024}+1\right)\left(5^{1024}-1\right)\left(5^{2048}+1\right)=\left(5^{2048}-1\right)\left(5^{2048}+1\right)=5^{4096}-1\)

\(\Rightarrow A=\frac{5^{4096}-1}{4}nha\)

Bài 1:

\(\left(3x+4\right)^2=9x^2+24x+16\)

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

\(\left(x-3\right)^2=x^2-6x+9\)

\(\left(\dfrac{1}{2}x-5\right)^2=\dfrac{1}{4}x^2-5x+25\)

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

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

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

\(4x-\dfrac{1}{9}=\left(2\sqrt{x}+\dfrac{1}{3}\right)\left(2\sqrt{x}-\dfrac{1}{3}\right)\)

Bài 3:

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

\(x^2-10x+25=\left(x-5\right)^2\)

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

\(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)

\(\left(\dfrac{2}{3}x+5\right)\left(\dfrac{2}{3}x-5\right)=\dfrac{4}{9}x^2-25\)

a. (x² + 4x + 4)

b. (1 - 4x + 4x²)

c. 9x² - 1

d. x³ + 9

e. x³ + 3x²4 + 19x + 64

f. x³ - 8

Sửa đề

B = 2(3+1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

= (3-1)(3+1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

= (3²-1)(3²+1)(3⁴+1)(3⁸+1)(3¹⁶+1)

= (3⁴-1)(3⁴+1)(3⁸+1)(3¹⁶+1)

= (3⁸-1)(3⁸+1)(3¹⁶+1)

= (3¹⁶-1)(3¹⁶+1)

= (3³²-1)