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a)
\(3^{n+1}+5.3^{n-2}=2592\)
\(\Rightarrow3^{n+1}+5.3^{n+1-3}=2592\)
\(\Rightarrow3^{n+1}+\dfrac{1}{27}.5.3^{n+1}=2592\)
\(\Rightarrow3^{n+1}+\dfrac{5}{27}.3^{n+1}=2592\)
\(\Rightarrow3^{n+1}.\left(\dfrac{5}{27}+1\right)=2592\)
\(\Rightarrow3^{n+1}.\dfrac{32}{27}=2592\)
\(\Rightarrow3^{n+1}=2187\)
\(\Rightarrow3^{n+1}=3^7\)
\(\Rightarrow n+1=7\)
\(\Rightarrow n=6\)
b)
\(3^{n+2}.5.3^{n-1}=864\)
\(\Rightarrow3^{n+2}+\dfrac{1}{27}.5.3^{n+2}=864\)
\(\Rightarrow3^{n+2}\left(\dfrac{5}{27}+1\right)=864\)
\(\Rightarrow3^{n+2}.\dfrac{32}{27}=864\)
\(\Rightarrow3^{n+2}=729\)
\(\Rightarrow3^{n+2}=3^6\)
\(\Rightarrow n+2=6\)
\(\Rightarrow n=4\)
a.
\(5^n+5^{n+2}=650\)
\(5^n\left(1+5^2\right)=650\)
\(5^n\left(1+25\right)=650\)
\(5^n\cdot26=650\)
\(5^n=650:26\)
\(5^n=25\)
\(5^n=5^2\Rightarrow n=2\)
b.
\(3^{n+3}+5\cdot3^n=864\)
\(3^n\left(3^3+5\right)=864\)
\(3^n\left(27+5\right)=864\)
\(3^n\cdot32=864\)
\(3^n=864:32\)
\(3^n=27\)
\(3^n=3^3\Rightarrow n=3\)
a) 5n + 5n+2 = 650
=> 5n + 5n . 52 = 650
=> 5n (1 + 52) = 650
=> 5n . 26 = 650
=> 5n = 25
=> n = 2
b) 3n+ 3 + 5.3n = 864
=> 3n . 33 + 5.3n = 864
=> 3n(33 + 5) = 864
=> 3n . 32 = 864
=> 3n = 27
=> n = 3
a,
5n + 5n + 2 = 650
=> 5n + 5n.52 = 650
=> 5n(1 + 52) = 650
=> 5n.26 = 650
=> 5n = 25
=> n = 2
a) 5n +5n+2 = 650
5n + 5n.52 = 650
5n.(1+25 ) = 650
5n.26= 650
5n = 25 = 52
=> n = 2
b) 3n+3 +5.3n = 864
3n.33 +5.3n = 864
3n.(33+5) = 864
3n.32 = 864
3n = 27 = 33
=> n = 3
các bài cn lại bn dựa vào mak lm nha!
\(A=1+3+3^2+3^3+...+3^{101}\)
\(3A=3+3^2+3^3+3^4+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+3^4+...+3^{101}\right)-\left(1+3+3^2+3^3+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\left(3^{101}-1\right):2\)
Thu gọn tổng sau:
A=1+3+32+33+...+3100
B= 2100-299-298-297-...-22-2
C= 3100-399+398-397-...+32-3+1
a)
\(\left(\frac{1}{3}\right)^n\cdot27^n=3^n\)
\(\Rightarrow\left(\frac{1}{3}\cdot27\right)^n=3^n\)
\(\Rightarrow9^n=3^n\)
\(\Rightarrow\left(3^2\right)^n=3^n\)
\(\Rightarrow3^{2n}=3^n\)
\(\Rightarrow2n=n\)
\(\Leftrightarrow n=0\)
Vậy \(n=0\)
d) Ta có:
\(6^{3-n}=216\)
\(\Rightarrow6^{3-n}=6^3\)
\(\Rightarrow3-n=3\)
\(\Rightarrow n=3-3\)
\(\Rightarrow n=0\)
Vậy \(n=0\)\(\text{ }\)
a, 5n+5n+2=650
=>5n+5n.52=650
=>5n(1+25)=650
=>5n.26=650
=>5n=25
=>5n=52
=>n=2
Vậy n=2
a)\(5^x+5^{x+2}=650\)(=)\(5x.\left(1+25\right)=650\)(=)\(5^x.26=650\)(=)\(5^x=25\)=>x=2
b)\(3^{x-1}+5.3^{x-1}=162\)(=)\(3^{x-1}.\left(1+5\right)=162\)(=)\(3^{x-1}.6=162\)(=)\(3^{x-1}=27\)(=)\(3^{x-1}=3^3\)=>x-1=3(=)x=2
c)\(4^x+4^{x+3}=4160\)(=)\(4^x.\left(1+64\right)=4160\)(=)\(4^x.65=4160\)(=)\(4^x=64\)(=)\(4^x=4^3\)
=>x=3
học tốt
\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
<=> \(3^{n-1}+5.3^{n-1}=162\)
<=> \(3^{n-1}\left(1+5\right)=162\)
<=> \(3^{n-1}.6=162\)
<=> \(3^{n-1}=162:6\)
<=> \(3^{n-1}=27\)
<=> \(3^{n-1}=3^3\)
<=> n - 1 = 3
<=> n = 3 + 1 = 4
Câu 1
a) Từ gt=>\(\hept{\begin{cases}x-5=1-3x\\x-5=3x-1\end{cases}}\)
<=>\(\hept{\begin{cases}4x=6\\2x=-4\end{cases}}\)
<=>\(\hept{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}\)
b) Ta có: \(\hept{\begin{cases}\left(3x-1\right)^{100}\ge0,\forall x\in R\\\left(2y+1\right)^{200}\ge0,\forall x\in R\end{cases}}\)
Kết hợp với đề bài => \(\hept{\begin{cases}3x-1=0\\2y+1=0\end{cases}}\)
=>\(\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{2}\end{cases}}\)
Bài 2
\(\frac{1}{3}.3^n+5.3^{n-1}=162\)
<=>\(3^{n-1}+5.3^{n-1}=162\)
<=>\(6.3^{n-1}=162\)
<=>\(3^{n-1}=27=3^3\)
<=>\(n-1=3\)
<=>\(n=4\)
\(a,5^x+5^{x+2}=650\Leftrightarrow5^x+5^x+5^2=650\Leftrightarrow5^x.26=650\Leftrightarrow5^x=5^2\Leftrightarrow x=2\) x=2
b,Với x=0 khi đó 3^0-1+5.3^0-1=2 (loại)
Với x=1 khi đó 3^1+5.3^1=18 (loại)
Với x=2 khi đó 5.3^x-1>16 (loại)
Vậy không có x thỏa mãn
a) n = 6.
b) n = 4.