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\(\frac{2}{3}.\frac{4}{7}=\frac{8}{21}\)
\(\frac{3}{11}.2=\frac{6}{11}\)
\(4.\frac{2}{7}=\frac{8}{7}\)
\(\frac{8}{21}:\frac{2}{3}=\frac{8}{21}.\frac{3}{2}=\frac{21}{2.21}=\frac{1}{2}\)
\(\frac{3}{7}.\frac{7}{3}=\frac{21}{21}=1\)
\(\frac{3}{7}:\frac{3}{7}=\frac{3}{7}.\frac{7}{3}=\frac{21}{21}=1\)
lỡ tay bấm gửi trả lời luôn
\(\frac{2}{3}.\frac{1}{6}.\frac{9}{11}=\frac{2.9}{18.11}=\frac{2.9}{2.9.11}=\frac{1}{11}\)
\(\frac{2.3.4}{2.3.4.5}=\frac{6.4}{6.4.5}=\frac{24}{24.5}=\frac{1}{5}\)
\(\frac{x}{7}x\frac{4}{9}x5=\frac{3x4x5}{7x9}\)
\(\frac{x}{7}x\frac{4}{9}x5=\frac{20}{21}\)
\(\frac{x}{7}x\frac{4}{9}=\frac{20}{21}:5\)
\(\frac{x}{7}x\frac{4}{9}=\frac{4}{21}\)
\(\frac{x}{7}\) \(=\frac{4}{21}:\frac{4}{9}\)
\(\frac{x}{7}\) \(=\frac{9}{21}\) \(=\frac{3}{7}\)
Vay x la 3
ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
b) 1/3+1/3^2+1/3^3+1/3^4+1/3^5 (goi tong bang M)
3M=1+1/3+1/3^2+1/3^3+1/3^4
3M-M=1-1/3^5
2M=242/243
M=242/243*1/2=121/243
Ta có:
\(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{1}{3}-\frac{1}{30}-\frac{1}{5}-\frac{1}{10}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10}{30}-\frac{1}{30}-\frac{6}{30}-\frac{3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(\left(\frac{10-1-6-3}{30}\right)\)
= \(\left(\frac{1}{21}+\frac{1}{210}+\frac{1}{2010}\right)\)\(\times\)\(0\)
= \(0\)
\(\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times\left(1-\frac{1}{6}\right)\times\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{4}\times\frac{1}{2}\)
\(=\frac{2}{8}-\frac{1}{8}=\frac{1}{8}\)
cái đuôi ak ko hiểu còn cái đầu thì dễ
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.......+\frac{1}{8.9.10}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+......+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+.......+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)