(x+5)3-x3-125
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\(x^3\) + 125 + (\(x\) + 5)(\(x\) - 25) = 0
(\(x^3\) + 53) + (\(x\) + 5)(\(x\) - 25) = 0
(\(x\) + 5)(\(x^2\) - 5\(x\) + 25) + (\(x\) + 5)(\(x\) - 25) =0
(\(x\) + 5)(\(x^2\) - 5\(x\) + 25 + \(x\) - 25) = 0
(\(x\) + 5)(\(x^2\) - 4\(x\)) = 0
\(x\)(\(x\) + 5)(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x=0\\x+5=0\\x-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\\x=4\end{matrix}\right.\)
a) 5^x=125
=> 5^x=5^2
=>x=2
b) 3^2*x=81
=>3^2*x=9^2
=>3^2*x=(3^2)^2
=>x=2
câu c chưa hiểu chỗ 5^2x3 hay 5^2*3 vậy ?
a) \(5^x=125\)
\(5^x=5^3\)
\(\Rightarrow x=3\)
b)\(3^{2x}=81\)
\(3^{2x}=3^4\)
\(\Rightarrow2x=4\)
\(x=4:2\)
\(x=2\)
c)\(5^{2x-3}-2.5^2=5^2.3\)
\(5^{2x-3}=5^2.3+5^2.2\)
\(5^{2x-3}=5^2.\left(3+2\right)\)
\(5^{2x-3}=5^3\)
\(\Rightarrow2x-3=3\)
\(2x=3+3\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
1/\(\left(x-6\right)^2=9\Rightarrow\orbr{\begin{cases}x-6=-3\\x-6=3\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=9\end{cases}}}\)
2/\(5^{x+1}=125\Leftrightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)
3/ đề thiếu
Bài 1:
a: \(\Leftrightarrow3^x\cdot10=810\)
\(\Leftrightarrow3^x=81\)
hay x=4
c: \(\Leftrightarrow5^x\cdot5+5^x\cdot\dfrac{1}{25}=126\)
\(\Leftrightarrow5^x\cdot\dfrac{126}{25}=126\)
\(\Leftrightarrow5^x=25\)
hay x=2
Bài 2:
a: \(27^{11}=3^{33}\)
\(81^8=3^{32}\)
mà 33>32
nên \(27^{11}>81^8\)
c: \(625^5=\left(5^4\right)^5=5^{20}\)
\(125^7=\left(5^3\right)^7=5^{21}\)
mà 20<21
nên \(625^5< 125^7\)
2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
(x+5)3-x3-125
=x3+15x2+75x+125-x3-125
=15x2+75x