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ta co : A= ( 8^9+12/8^9+7) -1
= 5/8^9+7
B=(8^10+4/8^10-1)-1
=5/8^10-1
VI 8^9+7 < 8^10-1 NEN 5/8^9+7 > 5/8^10-1
VAY A > B
Ta có : A = ( 8^9+12/8^9+7) - 1
= 5/8^9 + 7
B = (8^10+4/8^10-1) - 1
= 5/8^10-1
VI 8^9 + 7 < 8^10 - 1 nên 5/8^9+7 > 5/8^10-1
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~
a) Có \(\frac{n}{3n+1}=\frac{2n}{2\left(3n+1\right)}=\frac{2n}{6n+2}< \frac{2n}{6n+1}\)
=) \(\frac{n}{3n+1}< \frac{2n}{6n+1}\)
b) Có B < 1 =) \(B< \frac{10^8+1+9}{10^9+1+9}=\frac{10^8+10}{10^9+10}=\frac{10.\left(10^7+1\right)}{10.\left(10^8+1\right)}=\frac{10^7+1}{10^8+1}=A\)
=) B < A
lấy mik mặt cười ở đâu vậy nhắn tin mik nha mik kết bạn nha!!!!
a) Ta có: \(25^{15}=\left(5^2\right)^{15}=5^{30}\)
\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}\)\(=2^{30}.3^{30}=6^{30}\)
Vì \(5^{30}< 6^{30}\)nên \(25^{15}< 8^{10}.3^{30}\)
b) Ta có: \(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Vì \(2^{30}< 3^{30}\)nên \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)hay \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
_Học tốt_
a) 2515 và 810. 330
2515 = (52 ) 15 = 530
810. 330 = (23 )10. 330 = 230. 330 = 630
Vì 530< 630
nên 2515< 810. 330
b) \(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
\(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Vì \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)
nên \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
a)\(25^{15}=5^{2^{15}}=5^{30}\)
\(8^{10}.3^{30}=2^{3^{10}}.3^{30}=\left(2.3\right)^{30}=6^{30}\)
\(5^{30}< 6^{30}=>25^{15}< 8^{10}.3^{30}\)
b)\(\frac{4^{15}}{7^{30}}=\frac{2^{2^{15}}}{7^{30}}=\frac{2^{30}}{7^{30}}=\left(\frac{2}{7}\right)^{30}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{6^{30}}{14^{30}}=\left(\frac{6}{14}\right)^{30}=\left(\frac{3}{7}\right)^{30}\)
Vì hai số có mũ bằng 30 nên ta so sánh :\(\frac{2}{7}< \frac{3}{7}\)
=>\(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\).
Ta có : \(\frac{4^{15}}{7^{10}}=\frac{\left(2^2\right)^{15}}{7^{10}}=\frac{2^{30}}{7^{10}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.1^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}}=\frac{2^{30}.3^{30}}{7^{30}}=\frac{\left(2.3\right)^{30}}{7^{30}}=\frac{6^{30}}{7^{30}}\)
Mà : \(\frac{2^{30}}{7^{10}}=\frac{\left(2^3\right)^{10}}{7^{10}}=\frac{8^{10}}{7^{10}}\)
\(\frac{6^{30}}{7^{30}}=\frac{\left(6^3\right)^{10}}{\left(7^3\right)^{10}}=\frac{216^{10}}{343^{10}}\)
Vì : \(\frac{8}{7}>\frac{216}{343}\Rightarrow\frac{8^{10}}{7^{10}}>\frac{216^{10}}{343^{10}}\)
\(\Rightarrow\frac{4^{15}}{7^{10}}>\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
giúp mình vs
cho n là số tự nhiên
a, (n+ 10) (n+ 15) chia hết cho 2
b, n (n+ 1) (n+2) chia hết cho 2 và 3
c, n (n+ 1) (2n+1) chia hết cho 2 và 3
\(A=\frac{10^8+2}{10^8+1}=1+\frac{1}{10^8+1}<1+\frac{1}{10^8-3}<1+\frac{3}{10^8-3}=\frac{10^8}{10^8-3}=B\)
10^8+2>10^8
10^8+1<10^8-3