\(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\5x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
\(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
Với mọi giá trị của x ta có:
\(x^2\ge0\Rightarrow x^2+1\ge1\Rightarrow x=0\)
\(a,x+5x^2=0\\ \Leftrightarrow x\left(5x+1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\\\5x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy...
\(b,\left(x+1\right)=\left(x+1\right)^2\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy...
\(c,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow x=0\left(do...x^2+1>0\right)\)
Vậy...
a/ \(x+5x^2=0\)
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy ..
b/ \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x+1=\left(x+1\right)\left(x+1\right)\)
\(\Leftrightarrow x+1=x^2+2x+1\)
\(\Leftrightarrow x+1-x^2-2x-1=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy ....
a/ \(x^3+x=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy ..
1) x + 5x2 = 0
=>x(1+5x) = 0
=>\(\left[\begin{array}{} x=0\\ 1+5x=0 \end{array} \right.\)
=>\(\left[\begin{array}{} x=0\\ x=\dfrac{-1}{5} \end{array} \right .\)
Vậy .................
2)x+1 = (x+1)2
\(\Rightarrow\) (x+1) - (x+1)2 = 0
\(\Rightarrow\) (x+1)(1- x - 1) = 0
\(\Rightarrow\) -x(x+1) = 0
\(\Rightarrow\)\(\left[\begin{array}{} -x=0\\ x+1 = 0 \end{array} \right.\)
\(\Rightarrow\)\(\left[\begin{array}{} x = 0\\ x = -1 \end{array} \right.\)
Vậy ...................
3) x3 + x = 0
\(\Rightarrow\)x(x2 + 1) = 0
\(\Rightarrow\)\(\left[\begin{array}{} x = 0\\ x^{2} + 1 = 0 \end{array} \right. \)
\(\Rightarrow\)\(\left[\begin{array}{} x = 0\\ x \in \phi \end{array} \right.\)
\(\Rightarrow\) x = 0
Vậy ................
_______________JK ~ Liên Quân Group _____________
a,\(x+5x^2=0\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
b,\(x+1=\left(x+1\right)^2\Leftrightarrow x+1-\left(x+1\right)^2=0\Leftrightarrow\left(x+1\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c,\(x^3+x=0\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\)
x2+1=0 (vô lý, do x2+1>0)
Vậy x=0
CHÚC BẠN HỌC GIỎI...........
