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a, \(A=\frac{12}{3.7}+\frac{12}{7.11}+...+\frac{12}{195.199}\)
\(=3.\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{195.199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{195}-\frac{1}{199}\right)\)
\(=3.\left(\frac{1}{3}-\frac{1}{199}\right)\)
\(=3.\left(\frac{199}{597}-\frac{3}{597}\right)\)
\(=3.\frac{196}{597}\)
\(=\frac{196}{199}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{95.99}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)
\(B=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
Vậy giá trị của biểu thức \(B=\frac{32}{99}\)
Ta có : \(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+.....+\frac{4}{95.99}\)
\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{32}{99}\)
\(A=\frac{0,375-0,3+\frac{3}{10}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(\Rightarrow A=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(\Rightarrow A=\frac{-3.\left(-\frac{1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(-\frac{1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(\Rightarrow A=\frac{-3}{5}+\frac{3}{5}\)
\(\Rightarrow A=0\)
Vậy A = 0
@@ Học tốt @@
# Chiyuki Fujito
BÀI 1
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}.\)
bài 2
a) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
b) \(\frac{9^9.27^4}{3^8.81^5}=\frac{\left(3^2\right)^9.\left(3^3\right)^4}{3^8.\left(3^4\right)^5}=\frac{3^{18}.3^{12}}{3^8.3^{20}}=\frac{3^{30}}{3^{28}}=3^2=9\)
Study well
Bài 1: \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
Bài 2:
a)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{6-4+1}{12}=\frac{1}{4}\)
b)\(\frac{9^9.27^4}{3^8.81^5}=\frac{9^9.3^{12}}{3^8.9^{10}}=\frac{3^4}{9}=\frac{3^4}{3^2}=3^2=9\)
\(P=\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}\)
\(\Rightarrow P=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(\Rightarrow P=\frac{3}{5}\)
<=> \(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{3}\cdot\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}\cdot3=\frac{303}{1540}\)
<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
<=>\(x+3=308\)
<=>\(x=305\)
a) \(\frac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3.4^5}\)
= \(\frac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left(5^2\right)^3.\left(2^2\right)^5}\)
= \(\frac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{5^6.2^{10}}\)
= \(\frac{5^8.2^6-5^7.2^6+5^7.2^{10}}{5^6.2^{10}}\)
= \(\frac{5^7.2^6.\left(5-1+2^4\right)}{5^6.2^{10}}\)
= \(\frac{5.20}{2^4}=\frac{25}{4}\)
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}\)
\(=\frac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\frac{2^2}{5}=\frac{4}{5}\)
a) \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot13}+...+\frac{3}{647\cdot650}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}=\frac{1}{5}-\frac{1}{650}=\frac{129}{650}\)
b) \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)
\(=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)
sửa 199 -> 200
P/S : Lần sau đừng có đăng từng câu từng câu hỏi trên đây nhá
Bài giải
a, \(A=\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{647\cdot650}\)
\(A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{647}-\frac{1}{650}\)
\(A=\frac{1}{5}-\frac{1}{650}=\frac{13}{650}-\frac{1}{650}=\frac{12}{650}=\frac{6}{325}\)
b, \(B=\frac{12}{3\cdot7}+\frac{12}{7\cdot11}+...+\frac{12}{196\cdot200}\)
\(B=3\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+...+\frac{4}{196\cdot200}\right)\)
\(B=3\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{196}-\frac{1}{200}\right)\)
\(B=3\left(\frac{1}{3}-\frac{1}{200}\right)=3\cdot\frac{197}{600}=\frac{197}{200}\)