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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_
\(a)\)
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) 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) \(2\frac{7}{9}\)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
a) \(A=2^{24}=\left(2^3\right)^8=8^8.\)(1)
\(B=3^{16}=\left(3^2\right)^8=9^8\)(2)
Từ (1) và (2) \(\Rightarrow A< B\)
Vậy \(A< B.\)
b) \(B=\left(0,3\right)^{30}=\left(0,3^2\right)^{15}=0,09^{15}\)(1)
\(A=\left(0,1\right)^{15}\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
c) \(A=\left(\frac{-1}{4}\right)^8=\left(\frac{1}{4}\right)^8=\left[\left(\frac{1}{2}\right)^2\right]^8=\left(\frac{1}{2}\right)^{16}\)(1)
\(B=\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{2}\right)^3\right]^5=\left(\frac{1}{2}\right)^{15}\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
d) \(A=102^7=102^6.102\)(1)
\(B=9^{13}=9^{12}.9=\left(9^2\right)^6.9=81^6.9\)(2)'
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
e) \(8A=8\frac{8^{18}+1}{8^{19}+1}=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\)(1)
\(8B=8\frac{8^{23}+1}{8^{24+1}}=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)(2)
Từ (1) và (2) \(\Rightarrow8A>8B\Rightarrow A>B\)
Vậy \(A>B.\)
f) \(A=\frac{5^5}{5+5^2+5^3+5^4}=\frac{5^4}{1+5+5^2+5^3}=\frac{625}{156}>\frac{468}{156}=3.\)(1)
\(B=\frac{3^5}{3+3^2+3^3+3^4}=\frac{3^4}{1+3+3^2+3^3}=\frac{81}{40}< \frac{120}{40}=3.\)(2)
Từ (1) và (2) \(\Rightarrow A>B\)
Vậy \(A>B.\)
a, ta có A=2^24=64^4
B=3^16=81^4
Vì 64^4<81^4
Vậy 2^24<3^36
b, ta có A=0,1^15
B=0,3^30=0,09^15
Vì 0,1^15< 0,09^15
Vậy 0,1^15<0,3^30
a) Ta có 3/7 < 3,5/7 = 1/2 = 7,5/15 <11/15
Vậy 3/7 < 11/15
b) -11/6 < -1 < -8/9
Ai đủ điểm hỏi đáp đi ngang qua tk giùm với :(
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) Ta có : \(31^5< 32^5=\left(2^5\right)^5=2^{25}< 2^{28}=\left(2^4\right)^7=16^7< 17^7\)
\(\Rightarrow31^5< 17^7\)
b) Ta có : \(8^{12}=\left(2^3\right)^{12}=2^{36}>2^{32}=\left(2^4\right)^8=16^8>12^8\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)
\(2A=1-\frac{1}{99}\)
\(A=\frac{1-\frac{1}{99}}{2}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
a) \(31^5< 34^5=2^5.17^5=32.17^5\)
\(17^7=17^2.17^5=289.17^5\)
\(\Rightarrow31^5< 17^7\)
b) \(12^8< 16^8=\left(2^4\right)^8=2^{32}\)
\(8^{12}=\left(2^3\right)^{12}=2^{36}\)
\(\Rightarrow8^{12}>12^8\)
c) \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3A-A=1+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{3^2}-\frac{1}{3^2}\right)+...+\left(\frac{1}{3^{98}}-\frac{1}{3^{98}}\right)-\frac{1}{3^{99}}\)
\(\Rightarrow2A=1-\frac{1}{3^{99}}< 1\Rightarrow A< \frac{1}{2}\)