Ta có :

121.11ⁿ=1331

11ⁿ= 1331:121

11ⁿ=11

11=11¹

<=> n=1

121 . 11^n = 1331

         11^n = 1331 : 121

         11^n = 11

=> n = 11

121 : 11ⁿ = 1331

=> 11ⁿ = 161051

=> 11ⁿ = 11⁵

=> n = 5

\(121:11^n=1331\)

\(11^n=\frac{121}{1331}\)

\(11^n=\frac{1}{11}\)

\(11^n=11^{-1}\)

\(\Rightarrow n=-1\)

Vậy \(n=-1\)

Vậy \(n=-1\)

1)\(8.2^n=128\Rightarrow2^n=128:8\Rightarrow2^n=16\Rightarrow2^n=2^4\Rightarrow n=4\)

2)\(121.11^n=1331\Rightarrow11^n=1331:121\Rightarrow11^n=11\Rightarrow n=1\)

3)\(7^n:49=343\Rightarrow7^n:7^2=7^3\Rightarrow7^n=7^3.7^2\Rightarrow7^n=7^5\Rightarrow n=5\)

nhớ **** cho mình nhé

a) \(5^2=n\\ n=5.5\\ n=25\)

b) \(n^3=125\\ n^3=5.5.5=5^3\\ n=5\)

d) \(11^{n+1}=1331\\ 11^{n+1}=11.11.11=11^3\\ n+1=3\\ n=2\)

a) \(11^n=1331\)

\(\Rightarrow11^n=11^3\)

\(\Rightarrow n=3\)

b) \(n^3=125\)

\(\Rightarrow n^3=5^3\)

\(\Rightarrow n=5\)

c) \(5^4=n\)

\(\Rightarrow625=n\)

\(\Rightarrow n=625\)

d) \(\left(n+1^2\right)=9\)

\(\Rightarrow n+1=9\)

\(\Rightarrow n=9-1\)

\(\Rightarrow n=8\)

a) 11^n = 1331

⇒ 11^n = 11^3

⇔ n = 3

b) n^ 3 = 125

⇒ n^3 = 5^3

⇔ n = 5

c) 5^4 = n 

⇒ n = 625

d) ( n + 1^2 ) = 9

⇒ ( n + 1 ) = 9

⇒ n = 8 

(4n + 5) : 3 - 121 : 11 = 4

(4n + 5) : 3 - 11 = 4

(4n + 5) : 3 = 4 + 11

4n + 5) : 3 = 15

4n + 5 = 15 × 3

4n + 5 = 45

4n = 45 - 5

4n = 40

⇒n = 10

a ) \(\left(4x+5\right)\div3-121\div11=4\)

\(\left(4x+5\right)\div3-11=4\)

\(\left(4x+5\right)\div3=4+11\)

\(\left(4x+5\right)\div3=15\)

\(\left(4x+5\right)=15\cdot3\)

\(4x+5=45\)

\(4x=45-5\)

\(4x=40\)

\(x=10\)

(4x + 5) : 3 - 121 : 11 = 4

=> (4x + 5) : 3 - 11 = 4

=> (4x + 5) : 3 = 15

=> 4x + 5 = 45

=> 4x = 40

=> x = 10

b) 1 + 3 + 5 + ... + x = 1600

=>[(x - 1) : 2 + 1] . (x + 1) : 2 = 1600

=> \(\left(\frac{x}{2}-\frac{1}{2}+1\right).\frac{x+1}{2}=1600\)

=> \(\frac{x+1}{2}.\frac{x+1}{2}=1600\)

=> \(\left(\frac{x+1}{2}\right)^2=1600\)

=> \(\frac{x+1}{2}=40\)

=> x + 1 = 80

=> x = 79

(2n-1)² = 121

=> 2n-1 = 11

2n = 11+1

2n = 12

n = 12:2

n = 6

\(\left(2n-1\right)^2=121\)

\(\Leftrightarrow\left(2n-1\right)^2=\orbr{\begin{cases}11^2\\\left(-11\right)^2\end{cases}}\)

Do \(n\in N\)\(\Rightarrow\)\(\left(2n-1\right)^2=11^2\)

\(\Leftrightarrow2n-1=11\)

\(\Leftrightarrow2n=12\)

\(\Leftrightarrow n=6\)

a) 5⁴ = 625

b) n³ = 5³

\(\Rightarrow\)n = 5

c) 11ⁿ = 11³

^{\(\Rightarrow\)}n = 3

5⁴ = n 

=> n = 625

n³ = 125

=> n³ = 5³

=> n = 5

11ⁿ = 1331

=> 11ⁿ = 11³

=> n = 3