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A = 1 + 4 + 42 + ... + 499
4A = 4 + 42 + ... + 4100
4A - A = 4100 - 1
3A = 4100 - 1
=> 4100 - 1 + 1 = 4x
=> 4100 = 4x
=> x = 100
\(\Leftrightarrow2^{x+1}.3^y=4^x.3^x\)
\(\Leftrightarrow2^{x+1}.3^y=2^{2x}.3^x\)
\(\Leftrightarrow\frac{3^y}{3^x}=\frac{2^{2x}}{2^{x+1}}\)
\(\Leftrightarrow3^{y-x}=2^{x-1}\)
Nếu \(x>1\Rightarrow\) vế trái lẻ, vế phải chẵn pt vô nghiệm
\(\Rightarrow x=1\Rightarrow3^{y-1}=1\Rightarrow y=1\)
a) \(3^2.x+2^3.x=51\)
\(\Leftrightarrow x\left(3^2+2^3\right)=51\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\)
Vậy
b) \(6^2.2-\left(84-3^2.x\right):7=69\)
\(\Leftrightarrow\left(84-3^2.x\right):7=3\)
\(\Leftrightarrow84-3^2.x=21\)
\(\Leftrightarrow3^2.x=63\)
\(\Leftrightarrow x=7\)
Vậy
Bài 1:
\(a.\left(-356+57\right)-\left(27-356\right)=-356+57-27+356=\left(-356+356\right)+\left(57-27\right)=30\) \(b.125.\left(-24+24.225\right)=125.\left(-24+5400\right)=125.\left(-24\right)+125.5400=-3000+675000=672000\)
\(c.26.\left(-125\right)-125.\left(-36\right)=-125.\left(26-36\right)=-125.\left(-10\right)=1250\)
Bài 2:
\(a.\left(2x-4\right)^2=0\)
\(\Rightarrow2x-4=0\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(b.\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\)
Để (x+5) chia hết cho (x+3) thì 2 phải chia hết cho (x+3)
\(\Rightarrow x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(x+3=1\Rightarrow x=-2\)
\(x+3=-1\Rightarrow x=-4\)
\(x+3=2\Rightarrow x=-1\)
\(x+3=-2\Rightarrow x=-5\)
Vậy \(x\in\left\{-2;-4;-1;-5\right\}\)
Bài 2:
a)\(\left(2x-4\right)^2=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
b)\(\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\in Z\)
Suy ra \(2⋮x+3\Rightarrow x+3\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)
a) \(4^n=4096\Rightarrow4^n=4^6\Rightarrow n=6\)
b) \(5^n=15625\Rightarrow5^n=5^6\Rightarrow n=6\)
c) \(6^{n+3}=216\Rightarrow6^{n+3}=6^3\Rightarrow n+3=3\Rightarrow n=0\)
d) \(x^2=x^3\Rightarrow x^3-x^2=0\Rightarrow x^2\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
e) \(3^{x-1}=27\Rightarrow3^{x-1}=3^3\Rightarrow x-1=3\Rightarrow x=4\)
f) \(3^{x+1}=9\Rightarrow3^{x+1}=3^2\Rightarrow x+1=2\Rightarrow x=1\)
g) \(6^{x+1}=36\Rightarrow6^{x+1}=6^2\Rightarrow x+1=2\Rightarrow x=1\)
h) \(3^{2x+1}=27\Rightarrow3^{2x+1}=3^3\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\)
i) \(x^{50}=x\Rightarrow x^{50}-x=0\Rightarrow x\left(x^{49}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x^{49}=1=1^{49}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4n = 4096
4n = 212
n = 12
5n = 15625
5n = 56
n = 6
6n+3 = 216
6n+3 = 23.33
6n+3 = 63
n + 3 = 3
Ta có:
\(2^{3^{2^3}}=2^{3^8}=2^{6561}=2^{3.2187}=8^{2187}\)
\(3^{2^{3^2}}=3^{2^9}=3^{512}\)
Ta thấy \(8^{2187}>3^{512}\Rightarrow2^{3^{2^3}}>3^{2^{3^2}}\)
\(2^{3^{2^3}}=2^{3^8}=2^{6561}\)
\(3^{2^{3^2}}=3^{2^9}=3^{512}\)
Tới đây mk chịu để mk suy nghĩ đã!
\(\left(\frac{1}{3}\right)^{2x-1}-\frac{1}{3^2}=-\frac{2}{27}\)
=> \(\left(\frac{1}{3}\right)^{2x-1}=-\frac{2}{27}+\frac{1}{9}\)
=> \(\left(\frac{1}{3}\right)^{2x-1}=\frac{1}{27}\)
=> \(\left(\frac{1}{3}\right)^{2x-1}=\left(\frac{1}{3}\right)^3\)
=> 2x - 1 = 3
=> 2x = 3 + 1
=> 2x = 4
=> x = 4/2 = 2