\(\frac{5632}{12915}\)<\(\frac{1}{2}\)

nếu mún thì 1 k

còn ko mún thì 50 k

bé hơn nha bạn

Ta có: 1/9 + 1/10 < 1/8+1/8 = 1/4

1/41+1/42< 1/40+1/40=1/20

=> 1/5+1/9+1/10+1/41+1/42<1/5+1/4+1/20=1/2

Vậy 1/5+1/9+1/10+1/41!+1/42<1/2

Mình cũng đang cân người giúp câu dó nên ko trả lời được đâu !

Ta có : \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)

\(\Rightarrow S=\dfrac{1}{5}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\) (đpcm)

Ta có : \(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\)

\(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)

\(\Rightarrow S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)(đpcm)

đpcm là gj

\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}

\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)

\(< \frac{1}{5}+\frac{1}{8}+\frac{1}{8}+\frac{1}{40}+\frac{1}{40}\)

\(=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)