\(a.\)

Ta sẽ biến đổi biểu thức  \(B\)  quy về dạng có thể dùng được hằng đẳng thức  \(\left(x-y\right)\left(x+y\right)=x^2-y^2\), khi đó:

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

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

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

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

Vì  \(2^{16}>2^{26}-1\)  nên  \(2^{16}>\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

Vậy,  \(A>B\)

Tương tự với câu  \(b\)  kết hợp với phương pháp tách hạng tử, khi đó xuất hiện hằng đẳng thức mới và dễ dàng đơn giản hóa biểu thức \(A\). Ta có:

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

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

                                                                                \(=\frac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)=\frac{1}{2}\left(3^{128}-1\right)\)

Mặt khác, do  \(\frac{1}{2}<1\)  nên   \(\frac{1}{2}\left(3^{128}-1\right)<3^{128}-1\)

Vậy,  \(B>A\)

Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)

\(< =>A>B\)

a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)

\(\Rightarrow A< B\)

b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

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

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

...

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

\(\Rightarrow A>B\)

c,d tương tự

Ta có 

\(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1\)

Mà \(B=2012^2\)

Suy ra \(B>A\)

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

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

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

\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{22}+1\right)\left(3^{64}+1\right)\)

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

\(2A=3^{128}-1\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)

Vậy \(A< B\)

Chúc bạn học tốt !!!

A.(3²-1)=4.(3²-1)(3²+1)(3⁴+1)...(3⁶⁴+1)=4.(3⁴-1)(3⁴+1)...(3⁶⁴+1)=  ...  =4.(3¹²⁸-1)

<=>8A=4B <=>2A=B =>B>A

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

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

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

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

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

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

\(2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}< 3^{128}-1=B\)

Vậy \(A< B\)

Gộp nhóm 4 => A = -4 * 500+2001+2002-2003=0

B = 

a) tạm bỏ số 1 ra => có 2012 số hạng=> có 1006 cặp =(-1)

=> A=1+-(-1).1006=-1005

\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)

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

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

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

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

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

a)

 Ta có

a chia 5 dư 4

=> a=5k+4 ( k là số tự nhiên )

\(\Rightarrow a^2=\left(5k+4\right)^2=25k^2+40k+16\)

Vì 25k^2 chia hết cho 5

    40k chia hết cho 5

    16 chia 5 dư 1

=> đpcm

2) Ta có

\(12=\frac{5^2-1}{2}\)

Thay vào biểu thức ta có

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

\(\Rightarrow P=\frac{\left[\left(5^2\right)^2-1^2\right]\left[\left(5^2\right)^2+1^2\right]\left(5^8+1\right)}{2}\)

\(\Rightarrow P=\frac{\left[\left(5^4\right)^2-1^2\right]\left[\left(5^4\right)^2+1^2\right]}{2}\)

\(\Rightarrow P=\frac{5^{16}-1}{2}\)

3)

\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)

\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

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

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

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

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

Suy ra A < B

b) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1=B-1\)

Suy ra A < B

Phần a bạn nhân thêm ở A là (2-1) là ra hằng đẳng thức, cứ thế mà triển. (Kết quả: A<B)

Phần b: phân tích A, ta có:

2015.2017= (2016-1).(2016+1)= 2016^2 -1 <2016^2

Suy ra: A<B

\(VT=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)

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

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)

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