Tìm x biết
5x+25=125
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\(5x+5x^2=43x^3\\ \Rightarrow43x^3-5x^2-5x=0\\ \Rightarrow x\left(43x^2-5x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\43x^2-5x-5=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=25+4.5.43=885\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5+\sqrt{885}}{86}\\x=\dfrac{5-\sqrt{885}}{86}\end{matrix}\right.\)
\(x^3\) + 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
\(\left[{}\begin{matrix}x=0\\x+5=0\\x-4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-5\\x=4\end{matrix}\right.\)
\(\left(x+5\right)+\left(x+10\right)+...+\left(x+25\right)-125=150\)
\(\Leftrightarrow x+5+x+10+...+x+25-125=150\)
\(\Leftrightarrow5x-50=150\)
\(\Leftrightarrow5x=100\)
\(\Leftrightarrow x=\dfrac{100}{5}=20\)
Vậy \(x=20\)
\(x\times125-x\times25=37800\\ \Rightarrow x\times\left(125-25\right)=37800\\ \Rightarrow x\times100=37800\\ x=378\)
x + x + 25 = 125
ta có : 2x + 25 = 125
2x = 125 - 25
2x = 100
x = 50
k mik nha
x + x + 25 = 125
=>2x+25=125
=>2x=125-25
=>2x=100
=>x=100:2
=>x=50
k nha
5x + 25 = 125
5x = 125 - 25
5x = 100
x = 100 : 5
x = 20
Vậy x = 20
5x + 25 = 125
5x = 125 - 25
5x = 100
x = 20
vậy x = 20