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\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)
=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)
=>\(156\cdot5^x=44\cdot89-16=3900\)
=>\(5^x=\dfrac{3900}{156}=25\)
=>x=2
Lời giải:
$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$
$5^x(1+5+5^2+5^3)=88.89:2-16$
$5^x.156=3900$
$5^x=3900:156=25=5^2$
$\Rightarrow x=2$
5x+5x+1+5x+2=31
5x + 5x + 5x = 31 - 2 - 1
15x = 28
x= 28/15
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
2⁵ˣ⁺¹ - 2⁵ˣ = 32
2⁵ˣ.(2 - 1) = 2⁵
2⁵ˣ = 2⁵
5x = 5
x = 5 : 5
x = 1
\(2^{5x+1}-2^{5x}=32\)
\(\Rightarrow2^{5x+1}-2^{5x}=2^5\)
\(\Rightarrow2^{5x}\cdot2-2^{5x}\cdot1=2^5\)
\(\Rightarrow2^{5x}\cdot\left(2-1\right)=2^5\)
\(\Rightarrow2^{5x}\cdot1=2^5\)
\(\Rightarrow2^{5x}=2^5\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=\dfrac{5}{5}\)
\(\Rightarrow x=1\)
3\(x^2\).(5\(x\) + 1) + 6\(x^3\).(5\(x\) + 2) = 9\(x^3\) .(5\(x\) + 3)
15\(x^3\) + 3\(x^2\) + 30\(x^4\) + 12\(x^3\) = 45\(x^4\) + 27\(x^3\)
(15\(x^3\) + 12\(x^3\)) + 3\(x^2\) + 30\(x^4\) - 45\(x^4\) - 27\(x^3\) = 0
27\(x^3\) + 3\(x^2\) - 15\(x^4\) - 27\(x^3\) = 0
3\(x^2\) - 15\(x^4\) = 0
3\(x^2\).(1 - 5\(x^2\)) = 0
\(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)
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
\(5^x+5^{x+2}+5^{x+4}=3255\)
\(\Rightarrow5^x\left(1+5^2+5^4\right)=3255\)
\(\Rightarrow5^x.651=3255\)
\(\Rightarrow5^x=5\Rightarrow x=1\)
Bạn thử xem lại đề nhé, giữa 3 số này là dấu cộng hay dấu nhân.
Nếu là dấu cộng thì ta có:
Nếu là dấu nhân thì ta có:
\(\frac{5x+4}{2006}+\frac{5x+3}{2007}=\frac{5x+2}{2008}+\frac{5x+1}{2009}\)
\(\Leftrightarrow\frac{5x+4}{2006}+1+\frac{5x+3}{2007}+1=\frac{5x+2}{2008}+1+\frac{5x+1}{2009}+1\)
\(\Leftrightarrow\frac{5x+2010}{2006}+\frac{5x+2010}{2007}=\frac{5x+2010}{2008}+\frac{5x+2010}{2009}\)
\(\Leftrightarrow\left(5x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}\right)=\left(5x+2010\right)\left(\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Leftrightarrow5x+2010=0\)
\(\Leftrightarrow5x=-2010\)
\(\Leftrightarrow x=-402\)
\(5^x+5^{x+1}+5^{x+2}=31.5^6\)
\(\Rightarrow\)\(5^x.\left(1+5^1+5^2\right)=31.5^6\)
\(\Rightarrow\)\(5^x.31=31.5^6\)
\(\Rightarrow\)\(5^x=5^6\)
\(\Rightarrow\)\(x=6\)