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Ta có A = (32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = 8(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (34 - 1)(34 + 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (38 - 1)(38 + 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (316 - 1)(316 + 1)(332 + 1)(364 + 1)
=> 8A = (332 - 1)(332 + 1)(364 + 1)
=> 8A = (364 - 1)(364 + 1)
=> 8A = 3128 - 1 (1)
Đặt B = 3126
=> 8B = 3126 . 8 = 3126.(32 - 1) = 3128 - 3126 (2)
Từ (1)(2) => 8A > 8B
=> A > B
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 !
\(A=26^2-24^2=\left(26-24\right)\left(26+24\right)< \left(27-25\right)\left(27+25\right)=27^2-25^2=B\)
Vậy A < B.
Áp dụng hằng đẳng thức số 3, ta có:
A=\(\left(26-24\right)\left(26+24\right)=2.50=100\)
B=\(\left(27-25\right)\left(27+25\right)=2.52=104\)
Vậy: A<B
So sánh 2603 và 3402
2603=23.201=(23)201=8201
3402=32.201=(32)201=9201
Vì 8 < 9
Nên 8201 > 9201
Vậy 2603 < 3402
Ta có :
2603 = 23.201 = 8201
3402 = 32.201 = 9201
Vì 8201 < 9201 nên 2603 < 3402
\(E=163^2+74\times163+37^2=163^2+2\times163\times37+37^2=\left(163+37\right)^2=200^2\)
\(F=147^2-94\times147+47^2=147^2-2\times147\times47+47^2=\left(147-47\right)^2=100^2\)
\(\frac{E}{F}=\frac{200^2}{100^2}=\left(\frac{200}{100}\right)^2=2^2=4\)
\(E=4F\)
B=\(2^{16}-1\)
\(A=2+1.2^2+1.2^4+1.2^8+1\)\(=\left(2.2^2.2^4.2^8\right)+\left(1+1+1+1\right)\)\(=2^{15}+4\)
mà \(2^{16}>2^{15}\)=> A>B
áp dụng hằng đẳng thức \(A^2\) - \(B^2\) = (A - B).(A + B)
A= \(26^2\) - \(24^2\) = (26 - 24).(26 + 24) = 2.50
B= \(27^2\) - \(25^2\)=(27 - 25).(27 + 25) = 2.51
=> Vì 50 < 51 nên A < B
Ta có: 1262=126.126=(123+3).126=123.126+3.126=123.126+378
123.129= 123.(126+3)=123.126+123.3=123.126+369
Vì 378>369 nên 123.126+378>123.126+369
⇒ 1262>123.129 hay 123.129<1262