1.Tìm x thỏa mãn điều kiện:
a)
(x^2-2x+5)(x-2)=(x^2+x)(x-5)
<=> (x^3-2x^2 -2x^2 +4x+5x-10)=(x^3 -5x^2 +x^2 -5x)
<=> (x^3-4x^2 +9x-10)=(x^3 -4x^2 -5x)
<=>9x-10=-5x
14x=10
x=5/7

b.
(2x^2-3x+1)(x^2-5)-(x^2-x)(2x^2-x-10)=5
(2x^2-3x+1)(x^2-5)-(x^2-x)[2(x^2 -5)-x]=5
(2x^2-3x+1)(x^2-5)-2(x^2-x).(x^2 -5)+x(x^2-x)=5
(x^2-5)[(2x^2-3x+1)-2(x^2-x)]+x(x^2-x)=5
(x^2-5)(-x+1)+x^2(x -1)=5
(x-1)[(-x^2+5)+x^2]=5
(x-1)[ 5]=5
x-1= 1
x= 2

a, (x^2 - 2x + 5)(x - 2) = (x^2 + x)(x - 5)
<=> x^3 - 2x^2 + 5x - 2x^2 + 4x - 10 = x^3 + x^2 - 5x^2 - 5x
<=> x^3 - x^3 - 2x^2 - 2x^2- x^2 + 5x^2 + 5x + 4x + 5x - 10 = 0
<=> (x^3 - x^3) + (- 2x^2 - 2x^2- x^2 + 5x^2) + (5x + 4x + 5x) - 10 = 0
<=> 0 + 0 + 14x - 10 = 0
<=> 14x - 10 = 0
<=> 14x = 10
<=> x = 10 : 14
<=> x = 5/7
Vậy x = 5/7
b, (2x^2 - 3x + 1)(x^2 - 5) - (x^2 - x)(2x^2 - x - 10) = 5
<=> (2x^2 - 3x + 1).(x^2 - 5) - (x^2 - x).[2.(x^2 - 5) - x] = 5
<=> (2x^2 - 3x + 1)(x^2 - 5) - 2(x^2 - x).(x^2 - 5) + x(x^2 - x) = 5
<=> (x^2 - 5)[(2x^2 - 3x + 1) - 2(x^2 - x)] + x(x^2 - x) = 5
<=> (x^2 - 5)(- x + 1) + x^2(x - 1) = 5
<=> - (x^2 - 5)(x - 1) + x^2(x - 1) = 5
<=> (x - 1)(- x^2 + 5 + x^2) = 5
<=> (x - 1).5 = 5
<=> x - 1 = 1
<=> x = 1 + 1
<=> x = 2
Vậy x = 2