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B=1/4x (7+9) + 1/5x(9+11) + 1/6x(11+13) + 1/7x13
=4+2+4+\(\frac{13}{7}\)=10+\(\frac{13}{7}\)=\(\frac{83}{7}\)
a) \(\frac{7}{9}\)và \(\frac{11}{13}\)
Ta có : \(1-\frac{7}{9}=\frac{2}{9};1-\frac{11}{13}=\frac{2}{13}\)
Vì \(\frac{2}{9}>\frac{2}{13}\)nên \(\frac{7}{9}< \frac{11}{13}\).
b) \(\frac{125}{121}\)và \(\frac{413}{409}\)
Ta có : \(\frac{125}{121}-1=\frac{4}{121};\frac{413}{409}-1=\frac{4}{409}\)
Vì \(\frac{4}{121}>\frac{4}{409}\)nên \(\frac{125}{121}>\frac{413}{409}\)
c) \(\frac{8}{7}\)và \(\frac{4}{5}\)
Vì \(\frac{8}{7}>1;1>\frac{4}{5}\)nên \(\frac{8}{7}>\frac{4}{5}\)
d) \(\frac{46}{35}\)và \(\frac{41}{38}\)
Chon phân số trung gian là : \(\frac{46}{38}\)
Vì \(\frac{46}{35}>\frac{46}{38};\frac{46}{38}>\frac{41}{38}\)nên \(\frac{46}{35}>\frac{41}{38}\)
\(A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Leftrightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
\(A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+\frac{4}{7.9.11}+\frac{4}{9.11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(\Rightarrow A=\frac{1}{1.3}-\frac{1}{11.13}=\frac{1}{3}-\frac{1}{143}=\frac{140}{429}\)
a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)
b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)
c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)
Tử số chung là \(2^3.3^2.11=792\)
\(\dfrac{11}{13}=\dfrac{11x72}{13x72}=\dfrac{792}{936}\)
\(\dfrac{6}{8}=\dfrac{6x12x11}{8x12x11}=\dfrac{792}{1056}\)
\(\dfrac{4}{5}=\dfrac{4x18x11}{5x18x11}=\dfrac{792}{990}\)
\(\dfrac{8}{9}=\dfrac{8x9x11}{9x9x11}=\dfrac{792}{891}\)
\(\dfrac{9}{11}=\dfrac{9x8x11}{11x8x11}=\dfrac{792}{968}\)
11 = 1 \(\times\) 11; 6 = 2 \(\times\) 3; 8 = 2 \(\times\) 2 \(\times\) 2; 9 = 3 \(\times\) 3
Vậy tử số chung là: 11 \(\times\) 2 \(\times\)2 \(\times\) 2 \(\times\) 3 \(\times\) 3 = 792
\(\dfrac{11}{13}\) = \(\dfrac{11\times72}{13\times72}\) = \(\dfrac{792}{936}\)
\(\dfrac{6}{8}\) = \(\dfrac{6\times132}{8\times132}\) = \(\dfrac{792}{1056}\)
\(\dfrac{4}{5}\) = \(\dfrac{4\times198}{5\times198}\) = \(\dfrac{792}{990}\)
\(\dfrac{8}{9}\) = \(\dfrac{8\times99}{9\times99}\) = \(\dfrac{792}{891}\)
\(\dfrac{9}{11}\) = \(\dfrac{9\times88}{11\times88}\) = \(\dfrac{792}{968}\)
(4.x):3-121:11=4
(4.x):3-11=4
(4.x):3=4+11
(4.x):3=15
4x=15:3
4x=5
x=\(\frac{5}{4}\)
\(=>\left(4.x\right):3-11=4\)
\(=>\left(4.x\right):3=4+11=15\)
\(=>4.x=15.3=45\)
\(=>x=\frac{45}{4}=11.25\)
số cần tìm là :
11 + 11 = 4
12 + 12 = 9
13 + 13 = 14
đáp số : 14
\(\frac{4}{9\times11}+\frac{4}{11\times13}+...+\frac{4}{119\times121}\)
\(=2\times\left(\frac{2}{9\times11}+\frac{2}{11\times13}+...+\frac{2}{119\times121}\right)\)
\(=2\times\left(\frac{11-9}{9\times11}+\frac{13-11}{11\times13}+...+\frac{121-119}{119\times121}\right)\)
\(=2\times\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{119}-\frac{1}{121}\right)\)
\(=2\times\left(\frac{1}{9}-\frac{1}{121}\right)\)
\(=\frac{224}{1089}\)