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\(121:11^n=1331\)
\(11^n=\frac{121}{1331}\)
\(11^n=\frac{1}{11}\)
\(11^n=11^{-1}\)
\(\Rightarrow n=-1\)
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
(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
\(\left(2n-1\right)^2=121\)
\(\Leftrightarrow\left(2n-1\right)^2=\orbr{\begin{cases}11^2\\\left(-11\right)^2\end{cases}}\)
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
121.11n=1331
11n = 1331:121
11n = 11
=> n = 1