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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
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) 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!!!!
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) 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 :\(\frac{2}{7}< \frac{3}{7}\)
=>\(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\).
\(a)\)
Ta có :
\(1-\frac{2}{3}=\frac{1}{3};1-\frac{4}{5}=\frac{1}{5};1-\frac{7}{8}=\frac{1}{8};1-\frac{3}{4}=\frac{1}{4}\)
\(1-\frac{9}{10}=\frac{1}{10};1-\frac{8}{9}=\frac{1}{9};1-\frac{5}{6}=\frac{1}{6};1-\frac{6}{7}=\frac{1}{7}\)
Do \(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}>\frac{1}{7}>\frac{1}{8}>\frac{1}{9}>\frac{1}{10}\)
\(\Rightarrow1-\frac{1}{3}< 1-\frac{1}{4}< 1-\frac{1}{5}< 1-\frac{1}{6}< 1-\frac{1}{7}< 1-\frac{1}{8}< 1-\frac{1}{9}< 1-\frac{1}{10}\)
\(\Rightarrow\frac{2}{3}< \frac{3}{4}< \frac{4}{5}< \frac{5}{6}< \frac{6}{7}< \frac{7}{8}< \frac{8}{9}< \frac{9}{10}\)
Nếu \(\frac{a}{b}\)là 1 số thuộc dãy trên thì số tiếp theo là :
\(\frac{a+1}{b+1}\)
\(b)\)
Ta có :
\(a\left(a+2\right)=a^2+2a\)
\(b\left(a+1\right)=ab+b\)
Sorry , đến bước này mik chịu
~ Ủng hộ nhé
Phần b) Ý bạn là so sánh \(\frac{a}{b}\)và \(\frac{a+1}{b+2}\)
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
a) \(2\frac{7}{9}\)và \(8\frac{1}{3}\)
Ta có:
\(2\frac{7}{9}=\frac{25}{9}\)
\(8\frac{1}{3}=\frac{25}{3}=\frac{25.3}{3.3}=\frac{75}{9}\)
Vì \(\frac{25}{9}< \frac{75}{9}\)nên \(2\frac{7}{9}< 8\frac{1}{3}\)
b) \(\frac{12}{7}\)và \(\frac{48}{28}\)
Ta có:
\(\frac{48}{28}=\frac{48:4}{28:4}=\frac{12}{7}\)
Mà \(\frac{12}{7}=\frac{12}{7}\)nên \(\frac{12}{7}=\frac{48}{28}\)
c) \(\frac{2^9}{\left(4^3\right)^8+45}\)và \(\frac{5^2}{\left(2^4\right)^3.12}\)
Ta có:
\(\frac{2^9}{\left(4^3\right)^8+45}=\frac{\left(2^2\right).2^7}{\left(2^5\right)^8+45}=\frac{\left(2^2\right).2^7}{2^{40}+45}=\frac{2^{31}}{45}\)
Tương tự với phân số kia
Phần d tương tự nha