7ⁿ : 7⁴ = 49

=> 7^n-4 = 49 = 7²

=> n - 4 = 2

=> n = 6

a) 7ⁿ : 7⁴ = 49

=> 7^{n - 4} = 49

=> 7^{n - 4}  = 7²

=>   n - 4  = 2

=>   n       = 2 + 4

=>   n       = 6

b) 5.4ⁿ + 13 = 125

=> 5.4ⁿ        = 125 - 13

=> 5.4ⁿ        = 112

=>    4ⁿ         = 112 : 5

=>    4ⁿ          = 112/5

=>    n \(\in\varnothing\)

\(7^n\div7^4=49\)

\(7^{n-4}=7^2\)

\(\Rightarrow n-4=2\)

\(n=2+4\)

\(n=6\)

Vậy n = 6

=))

 (7.n)¹⁹⁹² =7¹⁹⁹².n¹⁹⁹²=7².7¹⁹⁹⁰.n¹⁹⁹²=49.7¹⁹⁹⁰.n¹⁹⁹² chia hết cho 49

a)\(\left(x2+7\right).\left(x2-49\right)< 0\)

\(\left(x2+7\right).\left(x2-49\right)< 0\) chứng tỏ hai vế \(\left(x2+7\right)\) và \(\left(x2-49\right)\) khác dấu nhau .

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)

Vì \(\left(x2+7\right)\) > \(\left(x2-49\right)\)

Nên ta có:

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\left(x+7\right)=0\\\left(x-49\right)=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=-7\\x=49\end{matrix}\right.\)

Vậy hai số nguyên đó là -7 và 49 .

Còn phần còn lại bạn làm tương tự nhé !

Câu 4:

\(4^{n+2}-4^{n-1}=252\\ \Leftrightarrow4^{n-1}.\left(4^3-1\right)=252\\ \Leftrightarrow4^{n-1}.63=252\\ \Leftrightarrow4^{n-1}=\dfrac{252}{63}=4=4^1\\ \Rightarrow n-1=1\\ \Rightarrow n=2\)

Câu 7:

\(\left(\dfrac{36}{49}\right)^4:\left(-\dfrac{6}{7}\right)^6=\left(\dfrac{6}{7}\right)^{2.4}:\left(\dfrac{6}{7}\right)^6=\left(\dfrac{6}{7}\right)^8:\left(\dfrac{6}{7}\right)^6=\left(\dfrac{6}{7}\right)^{8-6}=\left(\dfrac{6}{7}\right)^2=\dfrac{36}{49}\)

Câu 8:

\(13^{96}=\left(13^3\right)^{32}=2197^{32}\\ 15^{64}=\left(15^2\right)^{32}=225^{32}\\ Vì:2197^{32}>225^{32}\left(Do:2197>225\right)\Rightarrow13^{96}>15^{64}\)

7^x . 49 = 7²⁷

7^x . 7² = 7²⁷

7^x = 7²⁷ : 7²

7^x = 7²⁵

x = 25

Vậy x = 25

\(7^n=49=7^2=\left(-7\right)^2=>\hept{\begin{cases}n=7\\n=-7\end{cases}}\) ,mà n \(\in\) N nên n=7

\(2^n=32=2^5=>n=5\)

7ⁿ = 49

=> 7² = 49

=> n = 2

2ⁿ = 32

=> 2⁵ = 32

=> n = 5