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Bài 2:
a: \(\left(a-b-2\right)^2-\left(2a-2b\right)\left(a-b-2\right)+a^2-2ab+b^2\)
\(=\left(a-b\right)^2-4\left(a-b\right)+4+\left(a-b\right)^2-2\left(a-b\right)\left(a-b-2\right)\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left[\left(a-b\right)^2-2\left(a-b\right)\right]\)
\(=2\left(a-b\right)^2-4\left(a-b\right)+4-2\left(a-b\right)^2+4\left(a-b\right)\)
\(=4\)
b: \(\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)\cdot...\cdot\left(2^{256}+1\right)-1\)
\(=\left(2^{64}-1\right)\left(2^{64}+1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{128}-1\right)\left(2^{128}+1\right)\left(2^{256}+1\right)-1\)
\(=\left(2^{256}-1\right)\left(2^{256}+1\right)+1\)
\(=2^{512}-1+1=2^{512}\)
c: \(24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)-5^{64}\)
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)-5^{64}\)
=-1
nhiều quá bạn ạ
hay bạn tìm hiểu cách thức chung làm dạng bài tìm GTNN chứ như thế này thì làm lâu lắm
mik chỉ tìm hiểu đc đến câu I còn lại mik k hiểu lắm, bn có lm đc k, giúp mik vs
Giúp vs @@Phạm Hoàng GiangTrần Quốc LộcTrần Thị Hươnghattori heijiTRẦN MINH HOÀNGAn Nguyễn BáRibi Nkok NgokKien Nguyen
Trần Đăng NhấtHung nguyen
Sửa đề bài 1 : Rút gọn
a,\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right).........\left(2^{32}+1\right)-2^{64}\)
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
Giải:
a) \(A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)
\(\Leftrightarrow2A=2\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^8-1\right)\)
\(\Leftrightarrow2A=3^{16}-1\)
\(\Leftrightarrow A=\dfrac{3^{16}-1}{2}\)
Vậy ...
b) \(B=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=2.12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
...
\(\Leftrightarrow2B=\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=5^{64}-1\)
\(\Leftrightarrow B=\dfrac{5^{64}-1}{2}\)
Vậy ...
a) \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2\)
\(=16\)
b) \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(=\dfrac{1}{2}\left(3^{64}-1\right)\)
Giải:
1) B = 272 - 252 = (27 - 25)(27 + 25) = 20.52
Suy ra A<B, vì 202<20.52
2) D = 20032 - 1 = 20032 - 12 = (2003 - 1)(2003 + 1) = 2002.2004
Suy ra C = D.
3) Nhân (2-1) vào E, ta đươc: E = (2-1)(2+1)(22+1)(24+1)(28+1)(216+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào E, ta được kế quả:
E = 232-1
Suy ra E<F
4) Nhân (3-1) vào G, ta đươc: 2G = (3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào G, ta được kế quả:
2G = 332-1
Suy ra G = (332-1)/2
Mà (332-1)/2 < 332/2
Suy ra G<H
5)
Nhân 2 vào I, ta đươc: 2I = 2.12(52+1)(54+1)(58+1)...(532+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào I, ta được kế quả:
2I = 564-1
Suy ra I = (564-1)/2
Mà (564-1)/2 < 564-1
Suy ra I<K.
Chúc chị học tốt!