Tính nhanh:
a)45 . 94 - 2 . 69/ 210 . 38 + 68 . 20
b)2100 - 299 + 298 - 297 +...+ 22 - 2
c)(1+1/3+1/5+...+1/99) : (1/1.99+1/3.97+...+1/99.1)
(dấu chấm là dấu nhân nha). Cảm ơn mn
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a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)
\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}+\frac{1}{99}\)
\(=2-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
\(=2-\frac{1}{99}\)
\(=\frac{197}{99}\)
1+1/3+1/5...+1/97+1/99=(1+1/99) + (1/3+1/97) + (1/5+1/95)....+(1/49+1/51)
= 100/1.99 + 100/3.97 + 100/5.95 +.....=100.(1/1.99 + 1/3.97 + 1/5.95 +.....)
Mau so:
1/1.99 + 1/3.97 +1/5.95....+1/95.5+ 1/97.3 +1/99.1=2/1.99 +2/3.97 +2/5.95+.....
=2.(1/1.99 + 1/3.97 + 1/5.95 +.....)
=>A=(100.(1/1.99 + 1/3.97 + 1/5.95 +.....)) : (2.(1/1.99 + 1/3.97 + 1/5.95 +.....))=50
chuẩn luôn , tích nha
Thanks nhìu ^_^
Ta thấy:
1/1 + 1/99 = (99+1)/(1.99)=100/(1.99)
1/3 + 1/97 = (97+3)/(3.97)=100/(3.97)
1/5 + 1/95 = (95+5)/(5.95)=100/(3.97)
…
1/97 + 1/3 = (3+97)/(97.3)=100/(97.3)
1/99 + 1/1 = (1+99)/(99.1)=100/(99.1)
=>
1/(1.99)=(1/1+1/99)/100
1/(3.97)=(1/3+1/97)/100
…
1/(99.1)=(1/99+1/1)/100
------------------------------ cộng 2 vế của các đẳng thức trên. Ta được đẳng thức:
1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 )
=[(1/1+1/99)+(1/3+1/99)+…+(1/99+1/1)]/1...
=2(1+1/3+1/5+1/7…+1/99]/100
=(1+1/3+1/5+1/7…+1/99]/50
Vậy:
A=(1+1/3+1/5+1/7+...+1/97+1/99) / [ 1/(1.99) + 1/(3.97)+ 1/(5.95) +...+ 1/(97.3) + 1/(99.1 ) ]
A=(1+1/3+1/5+1/7+...+1/97+1/99)/[(1+1/3...
A=50.
\(A=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+...+2+1\)
\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)
\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)
\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)
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a) \(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)
b) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+...+3^{101}\)
\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)
\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)
c) \(C=5+5^2+...+5^{30}\)
\(\Rightarrow5C=5^2+5^3+...+5^{31}\)
\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)
\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)
d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
a)\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(3.2\right)^8.2^2.5}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+3^8.2^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+3^8.2^{10}.5}\)
\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
b) đặt A=2100 - 299 + 298 - 297 +...+ 22 - 2
=>2A=2101-2100+299-298+...+23-22
=>2A+A=2101-2100+299-298+...+23-22+2100 - 299 + 298 - 297 +...+ 22 - 2
=>3A=2101-2
=>A=\(\frac{2^{101}-2}{3}\)
thank kiu
thank kiu
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