A= $\frac{2}{1.5}$ + $\frac{3}{5.11}$ + $\frac{4}{11.19}$ + $\frac{5}{19.29}$ + $\frac{6}{29.41}$
B= $\frac{1}{1.4}$ + $\frac{2}{4.10}$ + $\frac{3}{10.19}$ + $\frac{4}{19.31}$
so sánh A và B
A= $\frac{2}{1.5}$ + $\frac{3}{5.11}$ + $\frac{4}{11.19}$ + $\frac{5}{19.29}$ + $\frac{6}{29.41}$ B= $\frac{1}{1.4}$ + $\frac{2}{4.10}$ + $\f
By Arianna
Ta có: A= $\frac{2}{1.5}$ + $\frac{3}{5.11}$ + $\frac{4}{11.19}$ + $\frac{5}{19.29}$ + $\frac{6}{29.41}$
⇒ A = $\frac{1}{2}$.(1-$\frac{1}{5}$ + $\frac{1}{5}$ – $\frac{1}{11}$ +…+ $\frac{1}{29}$ – $\frac{1}{41}$)
⇒ A = $\frac{1}{2}$.(1-$\frac{1}{41}$)
⇒ A = $\frac{1}{2}$ . $\frac{40}{41}$
⇒ A = $\frac{20}{41}$ (1)
Ta lại có: B = $\frac{1}{1.4}$ + $\frac{2}{4.10}$ + $\frac{3}{10.19}$ + $\frac{4}{19.31}$
⇒ 3B = $\frac{3}{1.4}$ + $\frac{6}{4.10}$ + $\frac{9}{10.19}$ + $\frac{12}{19.31}$
⇒ 3B = 1- $\frac{1}{4}$ + $\frac{1}{4}$ – $\frac{1}{10}$ + $\frac{1}{10}$ – $\frac{1}{19}$ + $\frac{1}{19}$ – $\frac{1}{31}$
⇒ 3B = 1- $\frac{1}{31}$
⇒ 3B = $\frac{30}{31}$
⇒ B = $\frac{10}{31}$ (2)
Từ (1) và (2) ⇒ A>B (Vì $\frac{20}{41}$ > $\frac{20}{62}$ = $\frac{10}{31}$)
Vậy A>B
Đáp án:
`A>B`
Giải thích các bước giải:
`A=2/1.5+3/5.11+4/11.19+5/19.29+6/29.41`
`=>2A=(2^2)/(1.5)+(3.2)/(5.11)+(4.2)/(11.19)+(5.2)/(19.29)+(6.2)/(29.41)`
`=>2A=4/1.5+6/5.11+8/11.19+10/19.29+12/29.41`
`=>2A=1-1/5+1/5-1/11+1/11-1/19+1/19-1/29+1/29-1/41`
`=>2A=1-1/41`
`=>2A=40/41`
`=>A=40/41:2`
`=>A=20/41`
`B=1/1.4+2/4.10+3/10.19+4/19.31`
`=>3B=(1.3)/1.4+(2.3)/4.10+(3^2)/(10.19)+(4.3)/(19.31)`
`=>3B=3/1.4+6/4.10+9/10.19+12/19.31`
`=>3B=1-1/4+1/4-1/10+1/10-1/19+1/19-1/31`
`=>3B=1-1/31`
`=>3B=30/31`
`=>B=30/31:3`
`=>B=10/31`
`=>A=20/41>20/62=10/31=B`
Vậy `A>B`.