Toán học Toán Lớp 6: chứng tỏ `B = 1/2^2+1/5^2+1/6^2+1/7^2 +1/8^2<1` 19 Tháng Mười Hai, 2022 By Quỳnh Toán Lớp 6: chứng tỏ B = 1/2^2+1/5^2+1/6^2+1/7^2 +1/8^2<1
Ta có: 1/2^2=1/(1.2)=1/1-1/2 1/5^2=1/(2.5)=1/2-1/5 1/6^2=1/(5.6)=1/5-1/6 1/7^2=1/(6.7)=1/6-1/7 1/8^2=1/(7.8)=1/7-1/8 Ta lại có: =>B=1/1-1/2+1/2-1/5+1/5-1/6+1/6-1/7+1/7-1/8 =>B=1/1-1/8 =>B=7/8<1 Vậy B<1 Trả lời
$B=\dfrac{1}{2^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}<\:1$ $=\dfrac{1}{4}+\dfrac{1}{25}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}$ $=\dfrac{176400}{705600}+\dfrac{28224}{705600}+\dfrac{19600}{705600}+\dfrac{14400}{705600}+\dfrac{11025}{705600}$ $=\dfrac{176400+28224+19600+14400+11025}{705600}$ $=\dfrac{249649}{705600}$ $=0.35381$ $0.35381<1$(thỏa mãn) Trả lời
Ta có:
1/2^2=1/(1.2)=1/1-1/2
1/5^2=1/(2.5)=1/2-1/5
1/6^2=1/(5.6)=1/5-1/6
1/7^2=1/(6.7)=1/6-1/7
1/8^2=1/(7.8)=1/7-1/8
Ta lại có:
=>B=1/1-1/2+1/2-1/5+1/5-1/6+1/6-1/7+1/7-1/8
=>B=1/1-1/8
=>B=7/8<1
Vậy B<1
$B=\dfrac{1}{2^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}<\:1$
$=\dfrac{1}{4}+\dfrac{1}{25}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}$
$=\dfrac{176400}{705600}+\dfrac{28224}{705600}+\dfrac{19600}{705600}+\dfrac{14400}{705600}+\dfrac{11025}{705600}$
$=\dfrac{176400+28224+19600+14400+11025}{705600}$
$=\dfrac{249649}{705600}$ $=0.35381$
$0.35381<1$(thỏa mãn)