n=7

k nha , hứa rùi

\(lim\frac{1+2\cdot3^n-7^n}{5^n+2\cdot7^n}\)

\(=lim\frac{\frac{1}{7^n}+\frac{6^n}{7^n}-1}{\frac{5^n}{7^n}+\frac{14^n}{7^n}}\)

\(=lim\frac{0+\left(\frac{6}{7}\right)^n-1}{\left(\frac{5}{7}\right)^n+2}=\frac{-1}{2}\)

1) \(7.4^x=7.4^3\Leftrightarrow4^x=4^3;x=3\)

2) \(\frac{3}{2.5^x}=\frac{3}{2.5^{12}}\Leftrightarrow5^x=5^{12};x=12\)

\(2^x=2.2^8=2^9;x=9\)

4) \(5.3^x=7.3^5-2.3^5\Leftrightarrow5.3^x=3^5.\left(7-2\right)\)

\(\Leftrightarrow3^5.x=3^5.5;x=5\)

2.3^x + 3^{x - 1} = 7 . (3² + 2 . 6²)

=> 2.3^x + 3^{x - 1} = 567

=> 7 . 3^{x - 1} = 567

=> 3^{x - 1} = 567 : 7 = 81

=> x - 1 = 4

=> x = 5

a)2*3^x+3^x-1=7(3²+2*6²)

2*3^x+3^x-1=7(9+72)=7*81

2*3^x+3^x/3=567

2*3^x+3^x*1/3=567

(2+1/3)*3^x=567

7/3*3^x=567

3^x=567:7/3

3^x=243=3⁵

=>x=5

b) mk ko hiểu đề mấy, cái chỗ 7^x+2 là nhân vs 2 ak

a) Đặt A = 1.2 + 2.3 + ........ + (n-1)n

3A = 1.2.3 + 2.3.(4-1) + .... + (n-1)n[(n+1)-(n-2)]

3A = 1.2.3 + 2.3.4 - 1.2.3 + .... + (n-1)n(n+1) - (n-2)(n-1)n

3A = (1.2.3 - 1.2..3) + ... + (n-1)n(n+1)

A = \(\frac{\left(n-1\right)n\left(n+1\right)}{3}\)

b) Đặt B = 1² + 2² + ..... + n²

B = 1(2 - 1) + 2(3 - 1) + ..... + n[(n + 1) - 1]

B = 1.2 + 2.3 + .......... + n(n + 1) - (1+2+3+....+n)

B = A -  \(\frac{n\left(n+1\right)}{2}\)