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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!
e) 3n+2 + 5.3n + 1 = 216
=> 3n . 32 + 5.3n . 31 = 216
=> 3n . 9 + 15.3n = 216
=> 3n ( 9 + 15) = 216
=> 3n . 24 = 216
=> 3n = 9
=> n = 2
f) 5n + 1 - 5n - 1 = 1254 . 23 . 37
=> 5n . 5 - 5n . 1/5 = 1254 . 23 . 37
=> 5n ( 5 - 1/5) = 1254 . 23 . 37
=> 5n . 24/5 = 1254 . 23 . 37
=> n không thỏa mãn
1. Ta có: \(x\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
=> \(x\left(6-x\right)^{2003}-\left(6-x\right)^{2003}=0\)
=> \(\left(6-x\right)^{2003}\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}\left(6-x\right)^{2003}=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}6-x=0\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=6\\x=1\end{cases}}\)
Bài 2. Ta có: (3x - 5)100 \(\ge\)0 \(\forall\)x
(2y + 1)100 \(\ge\)0 \(\forall\)y
=> (3x - 5)100 + (2y + 1)100 \(\ge\)0 \(\forall\)x;y
Dấu "=" xảy ra khi: \(\hept{\begin{cases}3x-5=0\\2y+1=0\end{cases}}\) => \(\hept{\begin{cases}3x=5\\2y=-1\end{cases}}\) => \(\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{1}{2}\end{cases}}\)
Vậy ...
a) n = 6.
b) n = 4.