Question

[le bui trung thanh]

tìm x 

a,x(x+2018)-2x+4036=0

b,x+5=2(x+5)^2

c,(x^2+1)(2x-1)+2x=1

d,x/3-x^2/4=0

Answer 1

[l҉o҉n҉g҉ d҉z҉]

a) x( x + 2018 ) - 2x - 4036 = 0 

<=> x( x + 2018 ) - 2( x + 2018 ) = 0

<=> ( x + 2018 )( x - 2 ) = 0

<=> \(\orbr{\begin{cases}x+2018=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2018\\x=2\end{cases}}\)

b) x + 5 = 2( x + 5 )²

<=> x + 5 = 2( x² + 10x + 25 )

<=> x + 5 = 2x² + 20x + 50

<=> 2x² + 20x + 50 - x - 5 = 0

<=> 2x² + 19x + 45 = 0

<=> 2x² + 10x + 9x + 45 = 0

<=> 2x( x + 5 ) + 9( x + 5 ) = 0

<=> ( x + 5 )( 2x + 9 ) = 0

<=> \(\orbr{\begin{cases}x+5=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{9}{2}\end{cases}}\)

c) ( x² + 1 )( 2x - 1 ) + 2x = 1

<=> 2x³ - x² + 4x - 1 - 1 = 0

<=> 2x³ - x² + 4x - 2 = 0

<=> x²( 2x - 1 ) + 2( 2x - 1 ) = 0

<=> ( 2x - 1 )( x² + 2 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}\Leftrightarrow}x=\frac{1}{2}\)( vì x² + 2 ≥ 2 > 0 ∀ x )

d) \(\frac{x}{3}-\frac{x^2}{4}=0\)

\(\Leftrightarrow\frac{4x}{12}-\frac{3x^2}{12}=0\)

\(\Leftrightarrow\frac{4x-3x^2}{12}=0\)

\(\Leftrightarrow4x-3x^2=0\)

\(\Leftrightarrow x\left(4-3x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)