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Đặt \(\hept{1\begin{cases}\frac{a_2}{a_1}=x\\\frac{b_2}{b_1}=y\\\frac{c_2}{c_1}=z\end{cases}}\)
Thì bài toán thành
x + y + z = 1(1); \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\left(2\right)\)
Chứng minh x2 + y2 + z2 = 1
Từ (2) ta có \(\frac{xy+yz+zx}{xyz}=0\Leftrightarrow xy+yz+zx=0\)
Từ (1) ta có
(x + y + z)2 = 1
<=> x2 + y2 + z2 + 2(xy + yz + zx) = 0
<=> x2 + y2 + z2 = 1
Bài 1:
a) \(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\)
\(=2^{16}-1\)
b) Sửa đề \(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)-3^{64}\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)-3^{64}\)
\(=3^{64}-1-3^{64}\)
\(=-1\)
Bài 2:
Ta có:
\(A=2009.2009\)
\(A=2009\left(2008+1\right)\)
\(A=2009.2008+2009\)
Ta lại có:
\(B=2008.2010\)
\(B=2008\left(2009+1\right)\)
\(B=2008.2009+2008\)
Vì 2008.2009 = 2009.2008
2009 > 2008
=> 2008.2009 + 2009 > 2009.2008 + 2008
=> A > B
1,a,(2-1)(2+1)(22+1)(24+1)(28+1)
=(22-1)(22+1)(24+1)(28+1)
=(24-1) (24+1)(28+1)
=(28 -1)(28+1)=216-1
2,
A=2009.2009=20092
B=2008.2010=(2009-1)(2009+1)=20092-1
Do20092>20092-1\(\Rightarrow A>B\)
O2 → Fe3O4 H2O → H2SO4 →Na2SO4
1)3Fe(bột) + 2O2 Fe3O4
Điều kiện: 150—600°C, cháy trong không khí
2) 4H2 + Fe3O4 3Fe + 4H2O
Điều kiện: trên 570°C
3) SO3 + H2O(lạnh) → H2SO4 (dung dịch pha loãng)
4Na2ZnO2 + 2H2SO4 →Na2SO4 + ZnSO4 + 2H2O
b) Fe2O3 + H2 → Fe + H2O
Fe + H2O → FeO + H2
2H2 + O2 →2H20
2H2O +2CL2 → 4HCL +O2
O2+2Ba→ 2BaO
BaO + H20→Ba(OH)2
b) A = 2010 . 2012
= ( 2011 - 1 )( 2011 + 1 )
= 20112 - 12 = 20112 - 1
20112 - 1 < 20112 => A < B
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}=B\)
a) Đặt \(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=2.\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^4-1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(...\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2A=3^{64}-1\)
\(A=\frac{3^{64}-1}{2}\)
Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1=A-1\)
Vậy B < A