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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
ĐẶT : A= \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)\(\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
= \(1-\frac{1}{99}=\frac{98}{99}\)
`@` `\text {Ans}`
`\downarrow`
`(1/2-1/3+1/4-1/5):(1/4-1/5)`
`=`\(\left(\dfrac{1}{6}+\dfrac{1}{20}\right)\div\dfrac{1}{20}\)
`=`\(\dfrac{1}{20}\div\dfrac{1}{20}+\dfrac{1}{6}\div\dfrac{1}{20}\)
`= 1+10/3`
`= 13/3`
A = (\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\)): (\(\dfrac{1}{4}\) - \(\dfrac{1}{5}\))
A = ( \(\dfrac{30}{60}\) - \(\dfrac{20}{60}\) + \(\dfrac{15}{60}\) - \(\dfrac{12}{60}\)):(\(\dfrac{5}{20}\) - \(\dfrac{4}{20}\))
A = \(\dfrac{13}{60}\): \(\dfrac{1}{20}\)
A = \(\dfrac{13}{60}\times\dfrac{20}{1}\)
A = \(\dfrac{13}{3}\)
\(1-2+3-4+5-6+.......+97-98+99-100+101\)
\(=\left(1-2\right)+\left(3-4\right)+\left(4-5\right)+.....+\left(97-98\right)+\left(99-100\right)+101\)
\(=50.\left(-1\right)+101=51\)
(5x6-30) = 0
0 nhân với mấy cũng bằng 0 . Vậy giá trị biểu thức là 0
a) \(\frac{2}{7}\div\frac{2}{3}-\frac{1}{7}\)
\(=\frac{2}{7}\times\frac{3}{2}-\frac{1}{7}\)
\(=\frac{2\times3}{7\times2}\)
\(=\frac{3}{7}-\frac{1}{7}\)
\(=\frac{2}{7}\)
b) \(\frac{2}{5}\times\frac{4}{3}\div\frac{2}{3}\)
\(=\frac{2}{5}\times\frac{4}{3}\times\frac{3}{2}\)
\(=\frac{2\times4\times3}{5\times3\times2}\)
\(=\frac{4}{5}\)
\(\dfrac{11}{2}\): \(\dfrac{1}{4}\) \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{11}{2}\) \(\times\) \(\dfrac{4}{1}\) \(\times\) \(\dfrac{5}{3}\)
= 22 \(\times\) \(\dfrac{5}{3}\)
= \(\dfrac{110}{3}\)
\(\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{5}{3}\)
= \(\dfrac{30}{12}-\dfrac{3}{12}+\dfrac{20}{12}\)
= \(\dfrac{7}{12}\)
\(\dfrac{14}{5}\times\dfrac{2}{3}\)+ 5
= \(\dfrac{28}{15}\) + 5
= \(\dfrac{28}{15}\) + \(\dfrac{75}{15}\)
= \(\dfrac{103}{15}\)
a: =1/2(3/4+1)=1/2x7/4=7/8
b: =9/8-1/6=27/24-4/24=23/24
\(A=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{4950}=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)