câu trr lơi kho qua

=796nha

=825 k mình nhé

Giúp me với

a) ( 1 - 1/2 ) x ( 1 - 1/3 ) x ( 1 - 1/4 ) x ( 1 - 1/5 ) x ( 1 - 1/6 )

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6 

= \(\frac{1.2.3.4.5}{2.3.4.5.6}\)

= 1/6

\(\frac{1}{6}\)

1/6

đáp số:1/6

a: \(\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\right):\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{1}{15}\right)\)

\(=\dfrac{5+3+2}{30}:\dfrac{5+3-2}{30}\)

\(=\dfrac{10}{30}\cdot\dfrac{30}{6}\)

\(=\dfrac{10}{6}=\dfrac{5}{3}\)

b: \(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\right):\left(\dfrac{1}{4}-\dfrac{1}{6}\right)\)

\(=\dfrac{30-20+15-12}{60}:\dfrac{3-2}{12}\)

\(=\dfrac{13}{60}\cdot\dfrac{12}{1}=\dfrac{13}{5}\)

\(\frac{\frac{1}{6}+\frac{1}{10}+\frac{1}{15}}{\frac{1}{6}+\frac{1}{10}-\frac{1}{15}}=\frac{\frac{5+3+2}{30}}{\frac{5+3-2}{30}}=\frac{\frac{1}{3}}{\frac{1}{5}}=1\frac{2}{3}\)

\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{1}{4}-\frac{1}{6}}=\frac{\frac{30-20+15-12}{60}}{\frac{3-2}{12}}=\frac{\frac{13}{60}}{\frac{1}{12}}=2\frac{3}{5}\)

tíc nha mn, mình đang bị âm điểm

thank you ban nha

2023 rồi còn ai ở đây ko ??? 😢😢😢

\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)

\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\frac{98}{99}\)

\(\Rightarrow B=\frac{98}{33}\)

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=1-\frac{1}{7}\)

\(\Rightarrow A=\frac{6}{7}\)

\(D=\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{128}\)

\(\Rightarrow2D=1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{64}\)

\(\Rightarrow2D-D=\left(1+\cdot\cdot\cdot+\frac{1}{64}\right)-\left(\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{128}\right)\)

\(\Rightarrow D=1-\frac{1}{128}\)

\(\Rightarrow D=\frac{127}{128}\)

\(a,\frac{6}{5\cdot7}+\frac{6}{7\cdot9}+\frac{6}{9\cdot11}+...+\frac{6}{103\cdot105}\)

\(=\frac{6}{2}\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+...+\frac{2}{103\cdot105}\right)\)

\(=\frac{6}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{103}-\frac{1}{105}\right)\)

\(=\frac{6}{2}\left(\frac{1}{5}-\frac{1}{105}\right)\)

\(=\frac{6}{2}\cdot\frac{20}{105}\)

\(=\frac{60}{105}\)

\(b,\left(1-\frac{1}{2}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)

\(=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{7}{8}\cdot\frac{9}{10}\)

\(=\frac{189}{640}\)

\(c,\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\left(1-\frac{1}{7}\right)\left(1-\frac{1}{9}\right)\)

\(=\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}\cdot\frac{8}{9}\)

\(=\frac{384}{945}\)