Tìm x biết:
a) 5^x+1 - 2.5^x = 75.
b)9^x+1 - 5.3^2x = 324.
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\(9^{x+1}+5.3^{2x}=324\)
\(9^x.9+5.\left(3^2\right)^x=324\)
\(9^x.9+5.9^x=324\)
\(9^x.\left(5+9\right)=324\)
\(9^x.14=324\)
\(9^x=\frac{324}{14}\)
\(\Rightarrow x\in\varnothing\)
\(9^{x+1}-5.3^{2x}=324=>9^x.9-5.\left(3^2\right)^x=324=>\left(3^2\right)^x.9-5.\left(3^2\right)^x=324\)
\(=>3^{2x}.\left(9-5\right)=324=>3^{2x}=\frac{324}{4}=81=3^4=>2x=4=>x=2\)
vậy x=2
tick nhé
a) \(8\left(x-2\right)=3\)
\(\Leftrightarrow x-2=\dfrac{3}{8}\)
\(\Leftrightarrow x=\dfrac{3}{8}+2\)
\(\Leftrightarrow x=\dfrac{19}{8}\)
Vậy \(x=\dfrac{19}{8}\)
b) \(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow9^x.9-\left(3^2\right)^x.5=324\)
\(\Rightarrow9^x.9-9^x.5=324\)
\(\Rightarrow9^x\left(9-5\right)=324\)
\(\Rightarrow9^x.4=324\)
\(\Rightarrow9^x=\dfrac{324}{4}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(9^{x+1}-5.3^{2x}=324\)
\(\Rightarrow9^x.9-5.9^x=324\)
\(\Rightarrow9^x\left(9-5\right)=324\)
\(\Rightarrow9^x.4=324\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
vậy x=2
\(\Rightarrow x=2\)
\(9^{x+1}-5.3^{2x}=324\)
\(9^x.9-5.9^x=324\)
\(9^x.\left(9-5\right)=324\)
\(9^x.4=324\)
\(9^x=81=9^2\)
\(\Rightarrow x=2\)
Ta có :
9x + 1 - 5 . 32x = 324
(32)x + 1 - 5 . 32x = 324
32x + 2 - 5 . 32x = 324
9 . 32x - 5 . 32x = 324
32x (9 - 5) = 324
⇒ 32x = 324 : 4 = 81 ⇒ x = 2
a) 5x + 1 - 2.5x = 75
<=> 5x.5 - 2.5x = 75
<=> 5x.3 = 75
<=> 5x = 25
<=> 5x = 52
<=> x = 2
Vậy x = 2
b) 9x + 1 - 5.32x = 324
<=> (32)x + 1 - 5.32x = 324
<=> 32x + 2 - 5.32x = 324
<=> 32x.32 - 5.32x = 324
<=> 32x . 4 = 324
<=> 32x = 81
<=> 32x = 34
<=> 2x = 4
<=> x = 2
Vậy x = 2
UIUYI