\(3^{-1}\cdot3^n+6\cdot3^{n-1}=7\cdot3^6\)
Tìm n biết:
\(3^{-1}\cdot3^n+6\cdot3^{n-1}=7\cdot3^6\)
Giúp mk nkoa! Còn 15'36 là hết tg rồi!
vào đây nha https://coccoc.com/search/math#query=+3%5E%E2%88%921%C2%B73%5En%2B6%C2%B73%5En%E2%88%921%3D7%C2%B736++
\(3^{-1}\cdot3^n+6\cdot3^{n-1}=7\cdot3^6\)
\(3^{n-1}+6\cdot3^{n-1}=7\cdot3^6\)
\(3^{n-1}\left(1+6\right)=7\cdot3^6\)
\(3^{n-1}\cdot7=7\cdot3^6\)
\(\Rightarrow3^{n-1}=3^6\)
\(\Rightarrow n-1=6\)
\(n=6+1=7\)
\(3^{-1}.3^n+6.3^{n-1}=7.3^6\)
\(\Leftrightarrow\)\(3^{n-1}.6=7.3^6\)
Tới đây bạn tự giải đi nhé.
P/s: Thi thì bạn phải tự giải nhé, bài dễ như vậy mà bạn cũng lười động não thì đắp mềm ngủ cho rồi
N=\(\frac{5\cdot\left(2^2\cdot3^2\right)\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^6}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot2^{18}}\)
Tinh N
tìm tất cả các số nguyên n thỏa mãn các đẳng thức sau
\(5^3\cdot25^n=5^{3n}\)
\(a^{\left(2n+6\right)\cdot\left(3n-9\right)}=1\)
\(\dfrac{1}{3}\cdot3^n=7\cdot3^2\cdot9^2-2\cdot3^n\)
a: \(5^3\cdot25^n=5^{3n}\)
\(\Leftrightarrow5^{3n}=5^3\cdot5^{2n}\)
=>3n=2n+3
hay n=3
b: \(a^{\left(2n+6\right)\left(3n-9\right)}=1\)
=>(2n+6)(3n-9)=0
=>n=-3 hoặc n=3
c: \(\dfrac{1}{3}\cdot3^n=7\cdot3^2\cdot3^4-2\cdot3^n\)
\(\Leftrightarrow3^n\cdot\dfrac{1}{3}+3^n\cdot2=7\cdot3^6\)
\(\Leftrightarrow3^n=3^7\)
hay n=7
\(\frac{2^5\cdot3^7-2^5\cdot3^6}{2^5\cdot3^6}\)
=\(\frac{2^5.3^6\left(3-1\right)}{2^5.3^6}=2\)
\(\frac{2^5.3^7-2^5.3^6}{2^5.3^6}=\frac{2^5.\left(3^7-3^6\right)}{2^5.3^6}=\frac{2^5.1458}{2^5.729}=\frac{1458}{729}=2\)
Vậy mới chuẩn nè:
\(\frac{2^5.3^7-2^5.3^6}{2^5.3^6}=\frac{2^5.3^6\left(3^1-1\right)}{2^5.3^6}=3-1=2\)
Bạn dùng t/c phân phối với rút gọn là được
1: \(\dfrac{\left(2^{12}\cdot3^5-4^6\cdot9^2\right)}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{\left(5^{10}\cdot7^3-25^5\cdot49^2\right)}{\left(125\cdot7\right)^3-5^9\cdot14^3}\)
2: Chứng Minh với \(\forall N\in Z\) thì B= \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
2:
\(B=3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\cdot9+3^n-2^n\cdot4-2^n\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10⋮10\)
\(\frac{2^{12}\cdot3^5-4^6\cdot3^6}{2^{12}\cdot9^3+8^4\cdot3^3}\)
\(\frac{2^{12}.3^5-\left(2^2\right)^6.3^6}{2^{12}.\left(3^2\right)^3+\left(2^3\right)^4.3^3}\)
\(\frac{2^{12}.3^5.\left(1-3^{ }\right)}{2^{12}.3^3.\left(3^3-1\right)}\)
\(\frac{2^{12}.3^5.\left(-2\right)}{2^{12}.3^3.8}\)
\(\frac{3^2.\left(-1\right)}{4}\)
\(\frac{-9}{4}\)
VẬy.......................
nhớ tk cho mình nha
Tính
\(\frac{5\cdot\left(2^2\cdot3^2\right)^9\cdot\left(2^2\right)^6-2\cdot\left(2^2\cdot3\right)^{14}\cdot3^4}{5\cdot2^{28}\cdot3^{18}+7\cdot2^{29}\cdot3^{18}}\)
\(\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}+7.2^{29}.3^{18}}\)
\(=\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{2^{28}.3^{18}.\left(5+7.2\right)}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}.19}=\frac{2^{28}.3^{18}.\left(5.4-2\right)}{2^{28}.3^{18}.19}\)
\(=\frac{5.4-2}{19}=\frac{18}{19}\)
1. a) Thực hiện phép tính:
\(A=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
b) Chứng minh rằng với mọi số nguyên dương n thì 3n + 2 - 2n + 2 + 3n - 2n chia hết cho 10.
\(a)A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\dfrac{2^{12}.3^5-\left(2^2\right)^63.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\dfrac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3\right)^3.7^3+5^9.\left(7.2\right)^3}\)
\(A=\dfrac{2^{12}.3^5-2^{12}.3^5}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.7^3.2^3}\)
\(A=\dfrac{0}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+5^3+2^3\right)}\)
\(A=0-\dfrac{5^4.\left(-6\right)}{1+125+8}\)
\(A=0-\dfrac{625.\left(-6\right)}{134}\)
\(A=\dfrac{-3750}{134}\)\(=\dfrac{-1875}{67}\)
\(b)3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.3^2-2^n.2^2+3^n-2^n\)
\(=(3^n.9+3^n)-\left(2^n.4+2^n\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\)
\(Suy\) \(ra:\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
b. Ta có: \(3^{n +2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=\left(3^n.3^2+3^n\right)-\left(2^{n-1}.2^3+2^{n-1}.2\right)\)
\(=3^n.\left(3^2+1\right)-2^{n-1}\left(2^3+2\right)\)
\(=3^n.10-2^{n-1}.10⋮10\)
Cho \(S_1-S_2+S_3-S_4+S_5=\frac{m}{n}\) với m, n nguyên tố cùng nhau. Biết:
\(S_1=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\)
\(S_2=\frac{1}{2\cdot3}+\frac{1}{2\cdot4}+\frac{1}{2\cdot5}+\frac{1}{2\cdot6}+\frac{1}{3\cdot4}+\frac{1}{3\cdot5}+\frac{1}{3\cdot6}+\frac{1}{4\cdot5}+\frac{1}{4\cdot6}+\frac{1}{5\cdot6}\)
\(S_3=\frac{1}{2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot5}+\frac{1}{2\cdot3\cdot6}+\frac{1}{2\cdot4\cdot5}+\frac{1}{2\cdot4\cdot6}+\frac{1}{2\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot6}+\frac{1}{3\cdot5\cdot6}+\frac{1}{4\cdot5\cdot6}\)
\(S_4=\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{2\cdot3\cdot4\cdot6}+\frac{1}{2\cdot3\cdot5\cdot6}+\frac{1}{2\cdot4\cdot5\cdot6}+\frac{1}{3\cdot4\cdot5\cdot6}\)
\(S_5=\frac{1}{2\cdot3\cdot4\cdot5\cdot6}\)
Tính \(m+n\)