Question

1. Tìm x:

a) x² - 3x = 0

b) 2x ( x - 2 ) - (x - 2 ) = 0

c) x + 5x² = 0

d) x + 1 = ( x + 1 )²

e) x³ + x = 0

f) 5 ( x + 3 ) - 2x ( 3 + x ) = 0

h) 6x ( x² - 2 ) - ( 2 - x² ) = 0

i) ( x + 1 )² - ( x + 1 ) ( x - 2 ) = 0

Answer 1

a)\(x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

b)2x ( x - 2 ) - (x - 2 ) = 0

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\)

c)\(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{5}\end{matrix}\right.\)

Answer 2

\(x^3+x=0\)

\(\Leftrightarrow x\left(x^2+1\right)=0\)

Vì \(x^2+1>0\forall x\)

nên x=0

Answer 3

f) 5 ( x + 3 ) - 2x ( 3 + x ) = 0

\(\Leftrightarrow\)5 ( x + 3 ) - 2x ( x +3 ) = 0

\(\Leftrightarrow\)( x + 3 ) . ( 5 - 2x ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{5}{2}\end{matrix}\right.\)

Answer 4

h) 6x ( x² - 2 ) - ( 2 - x² ) = 0

\(\Leftrightarrow6x\left(x^2-2\right)+\left(x^2-2\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\6x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\frac{1}{6}\end{matrix}\right.\)

Answer 5

i) ( x + 1 )² - ( x + 1 ) ( x - 2 ) = 0

\(\Leftrightarrow\left(x+1\right)\left(x+1-x+2\right)=0\)

\(\Leftrightarrow3\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)