tính tổng
A=1+3+32+33+34+...+3100
giúp em giải với ạ
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Bài 1:
a) \(\dfrac{65}{91}+\dfrac{-33}{55}=\dfrac{5}{7}+\dfrac{-3}{5}=\dfrac{25}{35}+\dfrac{-21}{35}=\dfrac{4}{35}\)
b) \(\dfrac{36}{-84}+\dfrac{100}{450}=\dfrac{-3}{7}+\dfrac{2}{9}=\dfrac{-27}{63}+\dfrac{14}{63}=\dfrac{-13}{63}\)
Ta có :
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(A=3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=3\left(\frac{1}{4}-\frac{1}{100}\right)\)
\(A=3.\frac{6}{25}\)
\(A=\frac{18}{25}\)
Vậy \(A=\frac{18}{25}\)
Chúc bạn học tốt ~
\(A=\frac{3}{4.5}+\frac{3}{5.6}+\frac{3}{6.7}+...+\frac{3}{99.100}\)
\(\Rightarrow A=3.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=3.\left(\frac{1}{4}-\frac{1}{100}\right)=\frac{3.24}{100}\)
\(=\frac{3.4.6}{25.4}\)
\(\Rightarrow A=\frac{18}{25}\)
Answer:
\(A=1-2+3-4+5-6+...+99-100\)
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(99-100\right)\)
\(=\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
Có cặp số:
\([\left(100-1\right):1+1]:2=50\) cặp
\(\Rightarrow A=\left(-1\right).50=-50\)
\(B=1+\left(-4\right)+2+\left(-5\right)+...+20+\left(-23\right)\)
\(=[1+\left(-4\right)]+[2+\left(-5\right)]+...+[20+\left(-23\right)]\)
\(=\left(-3\right)+\left(-3\right)+...+\left(-3\right)\)
Có cặp số:
\([\left(20-1\right):1+1]:2.2=20\) cặp
\(\Rightarrow B=\left(-3\right).20=-60\)
a, A = \(\dfrac{3^{10}\times10+3^{10}\times6}{3^9\times2^4}\)
A = \(\dfrac{3^{10}\times\left(10+6\right)}{3^9\times2^4}\)
A = \(\dfrac{3^{10}\times16}{3^9\times16}\)
A = 3
c, C = \(\dfrac{36^{10}\times25^{15}}{30^8}\)
C = \(\dfrac{\left(6^2\right)^{10}.\left(5^2\right)^{15}}{30^8}\)
C = \(\dfrac{6^{20}.5^{30}}{6^8.5^8}\)
C = 612.522
A = 1 + 3 + 32 + 33 + 34 +......+ 3100
=> 3A = 3 + 32 + 33 + 34 + 35 + ...... + 3101
=> 3A - A = (3 + 32 + 33 + 34 + 35 + ...... + 3101) - (1 + 3 + 32 + 33 + 34 +......+ 3100)
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
ko có dụ 34 thế đâu