Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
cho 2 tick nha:
8x+2x=25.2^2
->10x=25.4
->10x=100
->x=10
9^x-1=9
->9^x-1=9^1
->x-1=1
->x=2
câu c tương tự, bạn cho 2 tick lun nha
Ta có 24 mũ 3 : 12 mũ 3 = (24:12) mũ 3 = 2 mũ
25 mũ 6 : 25 mũ 4 = 25 mũ (4-2) = 25 mũ 2
81 mũ 2 : 9 mũ 2 = (81:9) mũ 2 = 9 mũ 2
32 mũ 4 : 2 mũ 6 = 16834
\(A=2+2^2+2^3+...+2^{61}+2^{62}+2^{63}\)
\(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{61}+2^{62}+2^{63}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{61}\left(1+2+2^2\right)\)
\(A=2.7+2^4.7+...+2^{61}.7\)
\(A=\left(2+2^4+...+2^{61}\right).7\Rightarrow A⋮7\)
Vậy ...
Ta có:
\(A=2+2^2+2^3+...+2^{63}\)
\(\Rightarrow A=\left(2+2^2+2^3\right)+...+\left(2^{61}+2^{62}+2^{63}\right)\)
\(\Rightarrow A=2\left(1+2+2^2\right)+...+2^{61}\left(1+2+2^2\right)\)
\(\Rightarrow A=2.7+...+2^{61}.7\)
\(\Rightarrow A=\left(2+...+2^{61}\right).7⋮7\)
\(\Rightarrow A⋮7\)
\(\Rightarrowđpcm\)
410 . 230
=410.210.3
=410.(23)10
=410.810
=(4.8)10
=3210
các câu còn lại chuyển về ntn là đc
4^10.2^30=(2^2)^10.2^30=2^20.2^30=2^50
9^25.27^4.81^3=(3^2)^25.(3^3)^4.(3^4)^3=3^50.3^12.3^12=3^74
25^50.125^5=(5^2)^50.(5^3)^5=5^100.5^15=5^115
64^3.4^8.16^4=(4^3)^3.4^8.(4^2)^4=4^9.4^8.4^8=4^25
\(a.4^{10}.2^{30}\Rightarrow\left(2^2\right)^{10}.2^{30}=2^{20}.2^{30}=2^{20+30}=2^{50}\)
\(\left(-2\right)^3.\left(-5\right)^2-41.\left(-8\right)\)
\(=\)\(-8.25-41.\left(-8\right)\)
\(=-8\left(25-41\right)\)
\(=-8.\left(-16\right)\)
\(=128\)
\(2018-1018:\left(4^2-3^2\right)\)
\(=2018-1018:7\)
\(=2018-\frac{1018}{7}\)
\(=\frac{13108}{7}\)
\(-5.\left(-25\right).\left(-50\right)+5.25.50\)
\(=-6250+6250\)
\(=0\)
\(C=\frac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+25^{26}+...+25^2+1}\)
\(C=\frac{25^{28}+25^{28}+25^{20}+...+25^4+1}{\left(25^{28}+25^{24}+25^{20}+...+25^4+1\right)+\left(25^{30}+25^{26}+...+25^2\right)}\)
\(C=\frac{25^{28}+25^{24}+25^{20}+...+25^4+1}{\left(25^{28}+25^{24}+...+25^4+1\right)+25^2\left(25^{28}+25^{24}+...+1\right)}\)
\(C=\frac{25^{28}+25^{24}+25^{20}+...+25^{24}+1}{\left(25^{28}+25^{24}+...+25^4+1\right)+\left(25^2+1\right)}\)
\(C=\frac{1}{25^2+1}\)
\(C=\frac{1}{626}\)