so sánh hai số bằng cách vận dụng hằng đẳng thức
a) A=\(2^{16}\) và B=\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
b) A=\(4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\) và B=\(3^{218}-1\)
Những câu hỏi liên quan
Bài 6. So sánh hai số bằng cách vận dụng hằng đẳng thức:a. A1999.2001và B2000^2b. A2^{16}và Bleft(2+1right)left(2^2+1right)left(2^4+1right)left(2^8+1right)c. A2011.2013và B2012^2d. A4left(3^2+1right)left(3^4+1right)....left(3^{64}+1right) và B3^{128}-1
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Bài 6. So sánh hai số bằng cách vận dụng hằng đẳng thức:
a. \(A=1999.2001\)và \(B=2000^2\)
b. \(A=2^{16}\)và \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
c. \(A=2011.2013\)và \(B=2012^2\)
d. \(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\) và \(B=3^{128}-1\)
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 \(B>A\)
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So sánh hai số bằng cách vận dụng hằng đẳng thức :
\(A=4.\left(3^2+1\right).\left(3^4+1\right)....\left(3^{64}+1\right)\) và
\(B=3^{128}-1\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{22}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\Rightarrow A=\frac{3^{128}-1}{2}< 3^{128}-1=B\)
Vậy \(A< B\)
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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
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So sánh hai số bằng cách vận dụng hằng đẳng thức :
a, A = 1999 . 2001 và B = \(2000^2\)
b, A = \(2^{16}\) và B = \(\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)
a)A=\(1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1\)
Vậy A < B
b) \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}=A\)
Vậy B < A
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a) Ta có: \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)\)
\(=2000^2-1^2< 2000^2\)
Vậy A < B.
b) Ta có: \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1< 2^{16}\)
Vậy A > B.
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\(A=1999.2001=\left(2000+1\right)\left(2000-1\right)=2000^2-1\)
\(A< B\)
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(B=\left(2^8-1\right)\left(2^8+1\right)\)
\(B=2^{16}-1\)
\(B< A\)
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So sánh hai số bằng cách vận dụng hằng đẳng thức :
A = \(4\cdot\left(3^2+1\right)\cdot\left(3^4+1\right).................\cdot\left(3^{64}+1\right)\)
và B = \(3^{128}-1\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)....\left(3^{64}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\left(3^{64}+1\right)\)
\(2A=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}< 3^{128}-1=B\)
Vậy \(A< B\)
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BT7: Tính
\(1,A=8\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(2,B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
1: A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)
=(3^8-1)(3^8+1)(3^16+1)
=(3^16-1)(3^16+1)
=3^32-1
2: B=(1-3^2)(1+3^2)*...*(1+3^16)
=(1-3^4)(1+3^4)(1+3^8)(1+3^16)
=1-3^32
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1
\(A=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(3^{16}-1\right)\left(3^{16}+1\right)\\ =3^{32}-1\)
\(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^2\right)\left(1+3^2\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^4\right)\left(1+3^4\right)\left(3^8+1\right)\left(3^{16}+1\right)\\ =\left(1-3^8\right)\left(1+3^8\right)\left(3^{16}+1\right)\\ =\left(1-3^{16}\right)\left(1+3^{16}\right)=1-3^{32}\)
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so sánh A và B
câu nào làm đc thì cảm ơn nhé <33 ko đc thì thôi bỏ đấy cũng đc
\(A=2^{16};;B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)...\left(3^{64}+1\right);;B=3^{138}-1\)
\(A=1000^2+1003^2+1005^2+1006^2;;B=1001^2+1002^2+1004^2+1007^2\)
1. Biết số tự nhiên a chia cho 5 dư 4. Chứng minh rằng \(a^2\) chia cho 5 dư 1
2. Rút gọn biểu thức : \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
3. Chứng minh hằng đẳng thức: \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{15}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\frac{1}{2}\left(5^{32}+1\right)=\frac{5^{32}+1}{2}\)
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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)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+c^2+3ab\left(a+b\right)+3\left(a+b\right)c\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ca+cb+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
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so sánh hai số
A=\(3^{32}-1\)
B= \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
Áp dụng liên tục a2 - b2 = (a - b)(a + b) để biến đổi . Ta có:
A = 332 - 1 = (316 - 1)(316 + 1) = (38- 1)(38 + 1)(316 + 1) = (34 - 1)(34 + 1)(38 + 1)(316 + 2) = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) =
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1) = 2.B
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Ta có 2B = \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
2B = (34-1)(34+1)(38+1)(316+1)
2B = (38-1)(38+1)(316+1)
Tương tự ta đc:
2B = 332-1
B= 332-1/2 hay B= A/2
Vậy A>B
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ta có : B=(3+1)(3^2 +1)(3^4 +1)(3^8 +1)(3^16 +1)
=>(3-1)B=(3-1)(3+1)(3^2 +1)(3^4 +1)(3^8 +1)(3^16 +1)
=>2B=((3^2 -1)(3^2 +1)(3^4 +1)(3^8 +1)(3^16 +1)
=>2B=(3^4 -1)(3^4 +1)(3^8 +1)(3^16 +1)
=>2B=(3^8 -1)(3^8 +1)(3^16 +1)
=.2B=(3^16 -1)(3^16 +1)
=>2B=3^32 -1
=>B=(3^32 -1)/2, mà A=3^32 -1 nên B=A/2 hay A=2B
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Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
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