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a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)
=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)
=>\(156\cdot5^x=44\cdot89-16=3900\)
=>\(5^x=\dfrac{3900}{156}=25\)
=>x=2
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Lời giải:
$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$
$5^x(1+5+5^2+5^3)=88.89:2-16$
$5^x.156=3900$
$5^x=3900:156=25=5^2$
$\Rightarrow x=2$
![](https://rs.olm.vn/images/avt/0.png?1311)
5x+5x+1+5x+2=31
5x + 5x + 5x = 31 - 2 - 1
15x = 28
x= 28/15
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a, 42x - 6 = 1
=> 42 x = 7
=> x = 6
b, 5x + 5x + 1 +5x + 2 = 775
=> 15 x + 3 = 775
=> 15 x = 772
=> x = 772/ 15
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2⁵ˣ⁺¹ - 2⁵ˣ = 32
2⁵ˣ.(2 - 1) = 2⁵
2⁵ˣ = 2⁵
5x = 5
x = 5 : 5
x = 1
\(2^{5x+1}-2^{5x}=32\)
\(\Rightarrow2^{5x+1}-2^{5x}=2^5\)
\(\Rightarrow2^{5x}\cdot2-2^{5x}\cdot1=2^5\)
\(\Rightarrow2^{5x}\cdot\left(2-1\right)=2^5\)
\(\Rightarrow2^{5x}\cdot1=2^5\)
\(\Rightarrow2^{5x}=2^5\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=\dfrac{5}{5}\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-3x\right)+\left(x-3\right)=\left(5x-1\right)-\left(5x+3\right)\)
\(\Rightarrow1-3x+x-3=5x-1-5x-3\)
\(\Rightarrow\left(1-3\right)-\left(3x-x\right)=\left(5x-5x\right)-\left(1+3\right)\)
\(\Rightarrow-2-2x=-4\)
\(\Rightarrow2x=-2+4\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
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5\(^{x+1}\) - 5\(^x\) = 2.28 + 8
5\(^x\).(5 - 1) = 520
5\(^x\).4 = 520
5\(^x\) = 520 : 4
5\(^x\) = 130
Với \(x\) = 0 ⇒ 5\(^x\) = 50 = 1 < 130 (loại)
Với \(x\) > 0 ⇒ 5\(^x\) = \(\overline{...5}\) \(\ne\) 130 (loại)
Vậy \(x\) \(\in\) \(\varnothing\)
\(5^{x+1}-5^x=2.2^8+8\\ 5^x\left(5-1\right)=512+8\\ 5^x.4=520\\ 5^x=\dfrac{520}{4}=130\)
Em xem lại đề
![](https://rs.olm.vn/images/avt/0.png?1311)
\(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\cdot5^1-5^x=500\)
\(\Rightarrow5^x\cdot5-5^x=500\)
\(\Rightarrow5^x\cdot\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=500:4\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Ta có: 5x+1=750-5x
\(\Leftrightarrow\)5x+1+5x=750
\(\Leftrightarrow\)5x.6=750
\(\Leftrightarrow\)5x=125
\(\Leftrightarrow\)5x=53
\(\Leftrightarrow\)x=3
Vậy .......
Hok tốt
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