Thu gọn M= \(\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
Giúp mình gấp với nhé!
Những câu hỏi liên quan
Thu gọn biểu thức sau :
a) \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot\left(\frac{1}{16}+1\right)\cdot\cdot\cdot\left(1+\frac{1}{2^{2n}}\right)\)
b) \(\left(2+1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)-2^{64}\)
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
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a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
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Tính
S=\(-1^2+2^2-3^2+4^2-...+2016^2\)
A=\(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}.4^{64}\)
Mình đang gấp giúp mình nha!
\(P=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
Giúp mình với mình cần gấp !!! Cảm ơn !
Rút gọn:
P = \(12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
Giúp mình với ạ!!!!
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(P=\frac{1}{2}\left(5^{32}-1\right)\)
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Thu gọn: \(3\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)\)
đặt 3(42 + 1 )(44 +1)......=A
5A=15(42+1)(44+1)......
=(42-1)(42+1)(44+1).....
=(44-1)(44+1).......
bạn giái tiếp là ra
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= 3[(42.44.48.416.432)+(1.1.1.1.1)]
= 3[432768 + 1]
= 1232768 + 3
Ko chắc nhé
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Rút gọn :
\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
Ta có: \(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right).\dfrac{1}{3}\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right).\dfrac{1}{3}\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right).\dfrac{1}{3}\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right).\dfrac{1}{3}\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right).\dfrac{1}{3}=\left(2^{64}-1\right).\dfrac{1}{3}=\dfrac{2^{64}-1}{3}\)
Vậy ...
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rút gọn
\(3.\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right).\left(2^{64}+1\right)\)
3 = 2^2 - 1
Áp dụng HĐT a^2 - b^2
kq : 2^128 - 1
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Rút gọn:
\(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(3(2^2+1)(2^4+1)(2^8+1)(2^16 +1) \)
\( = (2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)\)
\( = (2^4-1)(2^4+1)(2^8+1)(2^16+1) \)
\( = (2^8-1)(2^8+1)(2^16+1) \)
\(= (2^16 -1)(2^16+1) = 2^32 - 1\)
3(2^2 +1) (2^4 +1 ) (2^8 +1) (2^16 +1)
= (4-1)(2^2+1)(2^4 +1)(2^8+1)(2^16+1)
= [(2^2-1)(2^2+1)] (2^4+1) (2^8+1)(2^16+1)
=(2^4 -1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
= (2^16-1)(2^16+1)
= 2^23 -1
Chúc bạn học tốt
Rút gọn biểu thức:
a) \(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
b) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
a) (6x + 1)2 + (6x - 1)2 - 2(1 + 6x)(6x - 1)
= (6x + 1 - 6x + 1)2 = 4
b) 3(22 + 1)(24 + 1)(28 + 1)(216 +1)
= (22 - 1)(22 + 1)(24 + 1)(28 + 1)(216 + 1)
= (24 - 1)(24 + 1)(28 + 1)(216 + 1)
= (28 - 1)(28 + 1)(216 + 1)
= (216 - 1)(216 + 1) = 232 - 1
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