l chứng tỏ rằng B=`1/2^2`+`1/3^2`+`1/4^2`+`1/5^2+`1/6^2`+`1/7^2`+`1/8^2`<1 08/09/2021 Bởi Autumn l chứng tỏ rằng B=`1/2^2`+`1/3^2`+`1/4^2`+`1/5^2+`1/6^2`+`1/7^2`+`1/8^2`<1
Ta có: $\dfrac{1}{2^{2}}$ $<$ $\dfrac{1}{1.2}$ $\dfrac{1}{3^{2}}$ $<$ $\dfrac{1}{2.3}$ $\dfrac{1}{4^{2}}$ $<$ $\dfrac{1}{3.4}$ $\dfrac{1}{5^{2}}$ $<$ $\dfrac{1}{5.6}$ ……….. $\dfrac{1}{8^{2}}$ $<$ $\dfrac{1}{7.8}$ ⇒ $B$ $<$ $\dfrac{1}{1.2}$ $+$ $\dfrac{1}{2.3}$ $+$ ….+ $\dfrac{1}{7.8}$ ⇒ $B$ $<$ $1$ $-$ $\dfrac{1}{2}$ $+$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3}$ $+$…..+ $\dfrac{1}{7}$ $-$ $\dfrac{1}{8}$ ⇒ $B$ $<$ $1$ $-$ $\dfrac{1}{8}$ $<$ $1$ ( Vì $\dfrac{1}{8}$ $>$ $0$) Vậy $B$ $<$ $1$ Chúc me học tốt! Bình luận
Đáp án: Giải thích các bước giải: Ta có: $\dfrac{1}{2^{2}}< \dfrac{1}{1. 2}; \dfrac{1}{3^{2}}< \dfrac{1}{2. 3};…; \dfrac{1}{8^{2}}< \dfrac{1}{7.8}$ $⇒ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+….+ \dfrac{1}{8^{2}}< \dfrac{1}{2. 3};…; \dfrac{1}{8^{2}}< \dfrac{1}{7.8}$ $⇒ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+….+ \dfrac{1}{8^{2}}< 1- \dfrac{1}{8}< 1$ $⇒ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+….+ \dfrac{1}{8^{2}}< 1$ Bình luận
Ta có:
$\dfrac{1}{2^{2}}$ $<$ $\dfrac{1}{1.2}$
$\dfrac{1}{3^{2}}$ $<$ $\dfrac{1}{2.3}$
$\dfrac{1}{4^{2}}$ $<$ $\dfrac{1}{3.4}$
$\dfrac{1}{5^{2}}$ $<$ $\dfrac{1}{5.6}$
………..
$\dfrac{1}{8^{2}}$ $<$ $\dfrac{1}{7.8}$
⇒ $B$ $<$ $\dfrac{1}{1.2}$ $+$ $\dfrac{1}{2.3}$ $+$ ….+ $\dfrac{1}{7.8}$
⇒ $B$ $<$ $1$ $-$ $\dfrac{1}{2}$ $+$ $\dfrac{1}{2}$ $-$ $\dfrac{1}{3}$ $+$…..+ $\dfrac{1}{7}$ $-$ $\dfrac{1}{8}$
⇒ $B$ $<$ $1$ $-$ $\dfrac{1}{8}$ $<$ $1$ ( Vì $\dfrac{1}{8}$ $>$ $0$)
Vậy $B$ $<$ $1$
Chúc me học tốt!
Đáp án:
Giải thích các bước giải:
Ta có:
$\dfrac{1}{2^{2}}< \dfrac{1}{1. 2}; \dfrac{1}{3^{2}}< \dfrac{1}{2. 3};…; \dfrac{1}{8^{2}}< \dfrac{1}{7.8}$
$⇒ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+….+ \dfrac{1}{8^{2}}< \dfrac{1}{2. 3};…; \dfrac{1}{8^{2}}< \dfrac{1}{7.8}$
$⇒ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+….+ \dfrac{1}{8^{2}}< 1- \dfrac{1}{8}< 1$
$⇒ \dfrac{1}{2^{2}}+ \dfrac{1}{3^{2}}+….+ \dfrac{1}{8^{2}}< 1$