tìm x
5x+52x=650
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\(5^{2x+1}.5^{x+1}=5^{17}\)
\(\Rightarrow5^{2x+1+x+1}=5^{17}\)
\(\Rightarrow5^{3x+2}=5^{17}\)
\(\Rightarrow3x+2=17\)
\(\Rightarrow3x=15\Rightarrow x=5\)
a, x - 3 : 2 = 5 14 : 5 12
=> x - 3 : 2 = 5 2
=> x - 3 : 2 = 25
=> x – 3 = 25
=> x = 53
b, 30 : x - 7 = 15 19 : 15 18
=> 30 : x - 7 = 15
=> x – 7 = 2
=> x = 9
c, x 70 = x
=> x 70 - x = 0
=> x ( x 69 - 1 ) = 0
=>
d, 2 x + 1 3 = 9 . 81
=> 2 x + 1 3 = 9 3
=> 2x + 1 = 9
=> x = 4
e, 5 x + 5 x + 2 = 650
=> 5 x 1 + 5 2 = 650
=> 5 x . 26 = 650
=> 5 x = 25
=> x = 2
f, 4 x - 1 2 = 25 . 9
=> 4 x - 1 2 = 5 2 . 3 2
=> 4 x - 1 2 = 15 2
=> 4x – 1 = 15
=> x = 4
\(5^{2x+1}\cdot5^{2x+2}-5^x\cdot5^{3x+2}=100\)
\(\Leftrightarrow5^{4x+3}-5^{4x+2}=100\)
\(\Leftrightarrow625^x\left(5^3-5^2\right)=100\)
\(\Leftrightarrow625^x=1\)
hay x=0
\(\Rightarrow5^{2x+1+2x+2}-5^{x+3x+2}=100\\ \Rightarrow5^{4x+3}-5^{4x+2}=100\\ \Rightarrow5^{4x+2}\left(5-1\right)=100\\ \Rightarrow5^{4x+2}=25=5^2\\ \Rightarrow4x+2=2\Rightarrow x=0\)
Ta có: \(100< 5^{2x-1}\le5^6\)
\(\Leftrightarrow5^2< 5^{2x-1}\le5^6\)
\(\Leftrightarrow2x-1\in\left\{3;5\right\}\)
\(\Leftrightarrow2x\in\left\{4;6\right\}\)
hay \(x\in\left\{2;3\right\}\)
a). 5x = 125
=>5x = 53
=> x = 3
b) 32x = 81
=> 32x = 34
=> 2x = 4
=> x = 2
c). 52x-3 – 2.52 = 52 .3
=>52x: 53 = 52 .3 + 2.52
=>52x: 53 = 52 .5
=>52x = 52 .5.53
=>52x = 56
=> 2x = 6
=> x=3
a) 5x = 125
x = 125 : 5
x = 25
b) 32x = 81
x = 81 : 32
x = 81/ 32
c) 52x - 3 - 2.52 = 52.3
52x - 3 - 104 = 156
x - 3 - 104 = 156 : 52
x - 3 - 104 = 3
x - 3 = 3 + 104
x - 3 = 107
x = 107 + 3
x = 110
Ko chắc lắm , sai thì Sorry nhé .
Đặt t = 5x, ta có (1)⇔ 1/5.t2 + 5t = 250 ⇔ t2 + 25t - 1250 = 0
⇔ t = 25 hoặc t = -50(loại)
⇔ 5x ⇔ x = 2.
\(5^x+5^{2x}=650\\ \Rightarrow5^x+5^2.5^x=650\\ \Rightarrow5^x\left(1+25\right)=650\\ \Rightarrow5^x=25\\ \Rightarrow x=2\)
Vậy x = 2