a, (x-5)²+x(6-x)+15=0
x^2 - 50x + 25 + 6x - x^2 + 15 = 0
( x^2 - x^2 ) - 50x + 6x + 25 + 15 = 0
-44x + 40 = 0
-44x = -40
x = $\frac{-40}{-44}$
Vậy x = $\frac{10}{11}$
b, (x-2).(x²+2x+4)=19
x^3 - 8 = 19
x^3 = 19 + 8 = 27
=> x = 3
c, (x-1)²-(2x+3)²=0
( x - 1 - 2x - 3 ) ( x - 1 + 2x + 3 ) = 0
( -x - 4 ) ( 3x + 2) = 0
=> \(\left[ \begin{array}{l}-x - 4 = 0\\3x + 2=0\end{array} \right.\)
=>$\left \{ {{x=-4} \atop {x=-2/3}} \right.$
d) (x-1)³-x(x-2)²-(x-2)=0
x^3 - 3x^2 + 3x - 1^3 - x( x^2 - 4x + 4 ) - x + 2 = 0
x^3 - 3x^2 + 3x - 1^3 - x^3 + 4x^2 + 4x - x + 2 = 0
( x^3 - x^3 ) - ( 3x^2 - 4x^2) + ( 3x- x) - 1 + 2 = 0
x^2 + 2x + 1 = 0
(x + 1)^2 = 0
=> x + 1 = 0
=> x = -1
Vậy x = - 1
e) x(x+2)+x+2=0
x^2 + 2x + x + 2 = 0
x^2 + 2x + 1 + x + 1 = 0
( x + 1)^2 + ( x + 1) = 0
(x + 1 + 1)( x+ 1 ) = 0
( x + 2) ( x + 1) = 0
=> \(\left[ \begin{array}{l}x + 2 =0\\x + 1 = 0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=-2\\x=-1\end{array} \right.\)
Vậy x = { -2; -1}
