a) 52x . 5 = 3125
52x + 1 = 55
\(\Rightarrow\)2x + 1 = 5
\(\Rightarrow\)2x = 5 - 1 = 4
\(\Rightarrow\)x = 4 : 2 = 2
b) 5x + 5x + 3 = 630
5x + 5x . 53 = 630
5x . ( 1 + 53 ) = 630
5x . 126 = 630
5x = 630 : 126
5x = 5
\(\Rightarrow\)x = 5
c) 32x + 1 = 243
32x + 1 = 35
\(\Rightarrow\)2x + 1 = 5
\(\Rightarrow\)2x = 5 - 1 = 4
\(\Rightarrow\)x = 4 : 2 = 2
d) 32x + 1 = 9
32x + 1 = 32
\(\Rightarrow\)2x + 1 = 2
\(\Rightarrow\)2x = 2 - 1 = 1
\(\Rightarrow\)x = 1 : 2 = \(\frac{1}{2}\)
a) 5^2x.5=3135
=> 5^2x =3125:5
=> 5^2x =625
=> 5^2x = 5^4
=> 2x =4
=> x =4:2
=> x =2
Vậy x=2
_
chị chỉ gợi ý cko em cách làm được không ?
a) \(5^2x.5=3125\)
\(5^2.x.5=5^5\)
\(5^2.x=5^4\)
\(\Rightarrow x=5^4:5^2=5^2=25\)
b) \(5^x+5^x+3=630\)
\(5^x+5^x=630-3=627\)
\(5^x.2=627\)\(\Rightarrow5^x=627:2=313,5\)
KL: x không tồn tại.
Ai thấy đúng thì ủng hộ
a) \(5^{2x}\cdot5=3125\) c)\(3^{2x+1}=243\)
<=>\(5^{2x+1}=5^5\) <=>\(3^{2x+1}=3^5\)
<=>\(2x+1=5\) <=>\(2x+1=5\)
<=>\(2x=4\) <=>\(2x=4\)
<=>\(x=2\) <=>\(x=2\)
Vậy: \(x=2\) Vậy: \(x=2\)
b)\(5^x+5^{x+3}=630\) d)\(3^{2x+1}=9\)
<=>\(5^x+5^{x+3}=5+5^4\) <=>\(3^{2x+1}=3^2\)
=>\(x=1\) <=>\(2x+1=2\)
Vậy: \(x=1\) <=>\(2x=1\)
<=>\(x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
