Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{5}{2}+\frac{4}{11}+\frac{1}{11}+\frac{1}{30}+\frac{13}{60}=\frac{5}{2}+\frac{1}{30}+\frac{13}{60}+\frac{4}{11}+\frac{1}{11}\)
\(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}+\frac{5}{11}=\frac{33}{12}+\frac{5}{11}\)
\(\frac{363}{132}+\frac{60}{132}=\frac{423}{132}=\frac{36}{11}\)
a.. A = 310 × 11 + 310 × 5 / 39 × 24
A = 310 × (11 + 5) / 39 × 24
A = 310 × 16 / 39 × 24
A = 310 × 24 / 39 × 24
A = 3
C = 11 × 322 × 37 - 915 / (2 × 314)2
C = 11 × 329 - (32)15 / 22 × 328
C = 11 × 329 - 330 / 22 × 328
C = 329 × (11 - 3) / 22 × 328
C = 329 × 8 / 22 × 328
C = 329 × 23 / 22 × 328
C = 3 × 2
C = 6
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) \(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(1-\frac{1}{100}\right)\)
\(=2.\frac{99}{100}=\frac{99}{50}\)
\(A=\frac{7x\left(2x2\right)^5x3^{11}+2^{13}x\left(3x3\right)^5}{\left(2x3\right)^{10}+2^{12}x3^{10}}\)
\(A=\frac{7x2^{10}x3^{11}+2^{13}x3^{10}}{2^{10}x3^{10}+2^{12}x3^{10}}\)
tự làm tiếp
a)\(VT=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(=\frac{1}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right]\)
\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\)
\(=\frac{1}{2}-\frac{1}{3n+2}=\frac{3n+2}{2\cdot\left(3n+2\right)}-\frac{2}{2\cdot\left(3n+2\right)}\)
\(=\frac{3n+2-2}{6n+4}=\frac{3n}{6n+4}=VP\)