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\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)
\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)
a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)
=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )
=> \(14.\frac{27}{28}=\frac{419}{28}\)
b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)
=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)
bài 2 :
( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632
=> 211x + 22366 = 23632
=> 211x = 23632 - 22366
=> 211x = 1266
=> x = 1266 : 211
x = 6
Lời giải:
$\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}$
$< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}$
$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}$
1) \(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+...+\frac{11}{5^{12}}\)
\(5P=\frac{1}{5^1}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)
\(5P-P=\frac{1}{5^1}+\left(\frac{2}{5^2}-\frac{1}{5^2}\right)+\left(\frac{3}{5^3}-\frac{2}{5^3}\right)+...+\left(\frac{11}{5^{11}}-\frac{10}{5^{11}}\right)-\frac{11}{5^{12}}\)
\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}-\frac{11}{5^{12}}\)
Đặt \(A=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{11}}\)
\(5A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(5A-A=1+\frac{1}{5}-\frac{1}{5}+\frac{1}{5^2}-\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(4A=1-\frac{1}{5^{11}}\Rightarrow A=\frac{1-\frac{1}{5^{11}}}{4}\)
\(4P=\frac{1-\frac{1}{5^{11}}}{4}-\frac{11}{5^{12}}=\frac{1-\frac{1}{5^{11}}}{16}-\frac{11}{5^{12}\cdot4}< \frac{1}{16}\)
Bạn EᑕSTᗩSY ᗰᗩTᕼ ơi, \(a^{n^{n^{...}}}\)là lũy thừa tầng, lớp 6 nâng cao mới học nhé!
(-6,17 +3+5/9-2-36/97)*(1/3-1/4-1/12)=(-6,17+3+5/9-2-36/97)*(4/12-3/12-1/12)=(-6,17+3+5/9-2-36/97)*0=0
sao chỉ có mỗi dấu cộng với dấu trừ thế bạn ? đề còn đâu
\(\Rightarrow\frac{1}{a}=\frac{5}{6}-\frac{b}{3}=\frac{5-2b}{6}\Rightarrow6=a\left(5-2b\right)\)
Đến đây bạn lm tiếp nha