a) ( x +2 )2 - ( 3x - 1 ) ( x +2 ) = 0
<=> (x+2)(x+2-3x+1) = 0
<=> (x+2)(-2x+3) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\-2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}}\)
b) ( 2x - 1 )2 - ( 2x + 5 ) ( 2x - 5 ) = 18
<=> 4x2 -4x +1 - (4x2-25) =18
<=> 4x2 -4x +1 - 4x2 + 25 = 18
<=> - 4x + 26 = 18
<=> - 4x = 18 - 26
<=> - 4x = -8
<=> x = 2
c) ( 2x + 3 )2 - ( x - 5 )2 = 0
<=> [( 2x + 3 ) - ( x - 5 )].[( 2x + 3 ) + ( x - 5 )] = 0
<=> (2x +3 -x +5) . (2x +3 + x - 5) = 0
<=> (x +8)(3x-2) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+8=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-8\\x=\frac{2}{3}\end{cases}}}\)
d) 5x3 + 3x - 8 = 0
<=> (5x3 -5x) +(8x-8) = 0
<=> 5x(x2 - 1) + 8(x-1) = 0
<=> 5x(x - 1)(x+1) + 8(x-1) = 0
<=> (x - 1)[5x(x+1) + 8] = 0
<=> (x-1)(5x2+5x +8 ) = 0
<=> (x-1).5.(x2+x+8/5) = 0
<=> 5.(x-1)(x2+x+1/4 + 27/20) = 0
\(\Leftrightarrow\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{27}{20}\right]\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x+\frac{1}{2}\right)^2+\frac{27}{20}=0\end{cases}\Leftrightarrow x=1}\)vỉ \(\left(x+\frac{1}{2}\right)^2+\frac{27}{20}>0\)với mọi x
Vậy x = 1
