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= 1.2.3.....99/2.3.4....100
=1/100
k mk nha đáp án đúng đó
bài 1 :
\(\frac{2}{3}\)+\(\frac{1}{3}\)=\(\frac{3}{3}\)=1
\(\frac{3}{4}\)+\(\frac{2}{4}\)+\(\frac{1}{4}\)=\(\frac{4}{4}\)=1
\(\frac{4}{5}\)+\(\frac{3}{5}\)+\(\frac{2}{5}\)+\(\frac{1}{5}\)=\(\frac{10}{5}\)= 2
chúc bạn học tốt !!!
1. a) 3^2 .2^2 .2^4 = 3^2. 2^(2+4)=3^2. 2^6
b) 10^2. 10^3.10^5= 10^(2+3+5)= 10^10
c) x.x^5=x^(1+5)=x^6
d) a^3 .a^2 ,a^5= a^(3+2+5)= a^10
2. a) 2^3 và 3^2 . Ta có: 2^3 = 8, 3^2=9 => 2^3 < 3^2
b) 2^4 và 4^2. Ta có: 2^4= 16, 4^2=16 => 2^4 = 4^2
c) 2^5 và 5^2. Ta có: 2^5= 32, 5^2=25 => 2^5 > 5^2
d) 2^100 và 100. => 2^100 > 100
3. a) 3^8:3^4= 3^(8-4)=3^4
b) 10^8:10^2=10^(8-6)=10^2
c) a^6: a (a#0) = a^(6-1)=a^5
A=\(\frac{3.3}{8.11}\)+\(\frac{3.3}{11.14}\)+\(\frac{3.3}{14.17}\)+........+\(\frac{3.3}{197.200}\)
A=3\(\frac{3}{8.11}\)+3\(\frac{3}{11.14}\)+3\(\frac{3}{14.17}\)+............+3\(\frac{3}{197.200}\)
A=3.(\(\frac{3}{8.11}\)+\(\frac{3}{11.14}\)+\(\frac{3}{14.17}\)+..............+\(\frac{3}{197.200}\))
A=3.(\(\frac{1}{8}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{14}\)+\(\frac{1}{14}\)-\(\frac{1}{17}\)+.........+\(\frac{1}{197}\)-\(\frac{1}{200}\))
A=3.(\(\frac{1}{8}\)-\(\frac{1}{200}\))
A=3.(\(\frac{50}{400}\)-\(\frac{2}{200}\))
A=3.\(\frac{48}{400}\)
A=3.\(\frac{3}{25}\)
A=\(\frac{9}{25}\)
\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)
a) \(-\frac{8}{18}-\frac{15}{27}=-\frac{4}{9}-\frac{5}{9}=\frac{-9}{9}=-1\)
b) \(\frac{19}{24}-\left(-\frac{1}{2}+\frac{7}{24}\right)\)
\(=\frac{19}{24}+\frac{12}{24}-\frac{7}{24}=\frac{24}{24}=1\)
c) \(P=\frac{3^{11}.11+3^{11}.21}{3^9.2^5}\)
\(P=\frac{3^{11}.\left(11+21\right)}{2^9.2^5}=\frac{3^{11}.32}{2^9.32}=3^2=9\)
d) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
A=1+2^3+2^5+......+2^99
2^2A=2^2X(1+2^3+2^5+........+2^99)
4A= 2^2+2^5+2^7+...........+2^101
4A-A=(2^2+2^5+2^7+.............+2^101)-(1+2^3+2^5+........+2^99)
3A= 2^101+2^2-1
A=(2^101+2^2-1):3