`@` `\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 = \(\dfrac{13}{3}\)

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)

\(=1+\frac{1}{2}\cdot\frac{2.3}{2}+\frac{1}{3}\cdot\frac{3.4}{2}+...+\frac{1}{16}\cdot\frac{16.17}{2}\)

\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{17}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+...+\frac{17}{2}=\frac{1}{2}\left(2+3+...+17\right)=\frac{1}{2}\cdot\frac{16.19}{2}=4.19=76\)

B = 1 bạn nhé , đúng 100000000000% luôn

A =\(\left(1+\frac{2}{1}\right)\left(1+\frac{2}{2}\right)\left(1+\frac{2}{3}\right)\left(1+\frac{2}{4}\right)...\left(1+\frac{2}{26}\right)\left(1+\frac{2}{27}\right)\)

\(=\frac{3}{1}.\frac{4}{2}.\frac{5}{3}.\frac{6}{4}....\frac{28}{26}.\frac{29}{27}=\frac{28.29}{1.2}=14.29=406\)

a: \(A=\dfrac{2}{15}+\dfrac{13}{15}-\dfrac{1}{4}-\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{1}{2}\)

b: =5,4(-3,6-6,4)

=5,4*(-10)

=-54

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}\)

\(=\frac{1}{\frac{2.\left(2+1\right)}{2}}+\frac{1}{\frac{3.\left(3+1\right)}{2}}+\frac{1}{\frac{4.\left(4+1\right)}{2}}+.....+\frac{1}{\frac{50\left(50+1\right)}{2}}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{50.51}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{50}-\frac{1}{51}\right)\)

\(=2\left(\frac{1}{2}-\frac{1}{51}\right)=\frac{49}{51}\)