`1/4 +1/6 +1/8`

`= 6/24 + 4/24 + 3/24`

`=(6+4+3)/24`

`=13/24`

Ca này bó tay rồi, ghi rõ ra đc ko, khó hiểu quá

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(2A=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{20-18}{18.19.20}=\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}=\dfrac{1}{2}-\dfrac{1}{19.20}\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

Đặt \(A=\frac{1}{4^2}+\frac{1}{6^2}+\cdots+\frac{1}{\left(2n\right)^2}\)

\(=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)\)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{n^2}<\frac{1}{\left(n-1\right)\cdot n}=\frac{1}{n-1}-\frac{1}{n}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{n-1}-\frac{1}{n}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}<1-\frac{1}{n}<1\)

=>\(\frac14\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\right)<\frac14\)

=>\(A<\frac14\) (ĐPCM)

giúp mik với

\(\frac{6}{7}+\frac{2}{5}-\frac{1}{2}+2\frac{38}{91}-\frac{4}{5}=\frac{27300}{31850}+\frac{12740}{31850}-\frac{15925}{31850}+\) \(2\frac{13300}{31850}\)\(-\frac{25480}{31850}\)

\(=2\frac{27300+12740-15925+13300-25480}{31850}\)\(=2\frac{341}{910}\)

HỌC TỐT!!!

mn giúp e vs

cái này sai nha