Tìm x biết
a/ 3x+x2 =0 b/ 3x-1+5.3x-1=162
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\(a,5^x+5^{x+2}=650\\ \Rightarrow a,5^x+5^x.25=650\\ \Rightarrow26.5^x=650\\ \Rightarrow5^x=25\\ \Rightarrow5^x=5^2\\ \Rightarrow x=2\)
\(b,3^{x.1}+5.3^{x.1}=162\\ \Rightarrow3^x+5.3^x=162\\ \Rightarrow6.3^x=162\\ \Rightarrow3^x=27\\ \Rightarrow3^x=3^3\\ \Rightarrow x=3\)
\(b,3x+x^2=0\\ \Rightarrow x\left(3+x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ c,\left(x-1\right)\left(x-3\right)< 0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1< 0\\x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1>0\\x-3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 1\\x>3\left(vô.lí\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>1\\x< 3\end{matrix}\right.\end{matrix}\right.\)
Vậy 1<x<3
a, 3\(x\).(\(x\) - 1) + \(x\) - 1 = 0
(\(x\) - 1).(3\(x\) + 1) = 0
\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b, \(x^2\) - 6\(x\) = 0
\(x\).(\(x\) - 6) = 0
\(\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
a) 3x+x2=0
=> 3x+x.x=0
=> x(3+x)=0
=> x=0 hoặc 3+x=0
Vậy x thuộc {-3;0}
b) 3x-1+5.3x-1=162
3x-1(1+5)=162
3x-1.6=162
3x-1=162:6
3x-1=27
=>3x-1=33
=>x-1=3
=>x=2
Vậy x=2
Bạn Đinh Ngọc Duy Uyên làm sai phần b) rùi. kết quả là x=4 mà !!