Question

Giải các pt sau

a, \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)\)=0

b,\(\left(2x-1\right)\left(x-5\right)\left(x^2+3\right)\)=0

c,\(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)

d,\(\left(2x+3\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)

Answer 1

a); b) Do tích = 0 

=> Từng thừa số = 0 và ta nhận xét: \(x^2+2;x^2+3>0\)

=> a) \(\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

và câu b) \(\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)

Answer 2

a; *x-1=0 <=>x=1

    *2x+5=0 <=>x=-2,5

    *x²+2=0 <=> ko có x

b; tương tự a

Answer 3

a/ \(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)

Vì \(x^2\ge0\Rightarrow x^2+2\ge2>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)

Answer 4

c) 

pt <=> \(18x^2+12x+2=3x^2-5x-2\)

<=> \(15x^2+17x+4=0\)

<=> \(\left(3x+1\right)\left(5x+4\right)=0\)

<=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-\frac{4}{5}\end{cases}}\)

d) 

pt <=> \(2x^2-5x-12=-x^2+9x-20\)

<=> \(3x^2-14x+8=0\)

<=> \(\left(x-4\right)\left(3x-2\right)=0\)

<=> \(\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}}\)

Answer 5

a) ( x - 1 )( 2x + 5 )( x² + 2 ) = 0

<=> \(\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)( do \(x^2+2\ge2>0\forall x\))

b) ( 2x - 1 )( x - 5 )( x² + 3 ) = 0

<=> \(\orbr{\begin{cases}2x-1=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)( do \(x^2+3\ge3>0\forall x\))

c) 2( 9x² + 6x + 1 ) = ( 3x + 1 )( x - 2 ) 

<=> 18x² + 12x + 2 = 3x² - 5x - 2

<=> 18x² + 12x + 2 - 3x² + 5x + 2 = 0

<=> 15x² + 17x + 4 = 0

<=> 15x² + 5x + 12x + 4 = 0

<=> 5x( 3x + 1 ) + 4( 3x + 1 ) = 0

<=> ( 5x + 4 )( 3x + 1 ) = 0

<=> \(\orbr{\begin{cases}5x+4=0\\3x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=-\frac{1}{3}\end{cases}}\)

d) ( 2x + 3 )( x - 4 ) = ( x - 5 )( 4 - x )

<=> 2x² - 5x - 12 = -x² + 9x - 20

<=> 2x² - 5x - 12 + x² - 9x + 20 = 0

<=> 3x² - 14x + 8 = 0

<=> 3x² - 12x - 2x + 8 = 0

<=> 3x( x - 4 ) - 2( x - 4 ) = 0

<=> ( 3x - 2 )( x - 4 ) = 0

<=> \(\orbr{\begin{cases}3x-2=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=4\end{cases}}\)

Answer 6

a,\(\left(x-1\right)\left(2x+5\right)\left(x^2+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}}\)

\(\Leftrightarrow x^2+2\Leftrightarrow x^2=-2\)( vô lí )

b, \(\left(2x-1\right)\left(x-5\right)\left(x^2+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=5\end{cases}}\)\(\Leftrightarrow x^2+3\Leftrightarrow x^2=-3\)( vô lí )