b)(8x - 3)(3x - 2) - (4x + 7)(x + 4)=(2x + 1)(5x - 1)
<=> 24x^2-25x+6-4x^2-23x-28=10x^2+3x-1
<=>10x^2-51x-21=0
<=>x=(51+ căn3441)/20 hoặc x=(51-căn3441)/20

a,(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
(x-4)(2x+3-3x+5)+(x-5)(x-2)=0
(x-4)(-x+8)+(x-5)(x-2)=0
-x^2+12x-8.4+x^2-7x+10=0
5x=22; x=22/5

b,(8x-3)(3x+2)-(4x-7)(x+4)=(2x+1)(5x-1)
24x^2+7x-6-(4x^2+9x-28)=10x^2+3x-1
10x^2 -5x+23=0
<=>(x-5/2)^2 +9x^2+23-25/4>0
vo nghiem

c,2x^2+3(x-1)(x+1)=0
2x^2+3x^2-9=0
5x°2=9
x=±3/√5

a,(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
<=> 2x^2 - 8x + 3x - 12 + x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20
<=> 2x^2 + x^2 - 8x + 3x - 2x - 5x - 12 + 10 - 3x^2 + 12x + 5x - 20 = 0
<=> 2x^2 + x^2 -3x^2 - 8x + 3x - 2x - 5x + 12x + 5x - 12 + 10 - 20 = 0
<=> 5x - 22 = 0
<=> 5x = 22
<=> x = 22/5
Vậy x = 22/5

a,(2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> 2x^2 - 8x + 3x - 12 - 2x + x^2 - 5x + 10 = 3x^2 - 12x - 5x + 20
<=> 2x^2 - 8x + 3x - 12 - 2x + x^2 - 5x + 10 - 3x^2 + 12x + 5x - 20 = 0
<=> (2x^2 + x^2 - 3x^2) + (3x - 8x - 2x - 5x + 12x + 5x) - (12 - 10 + 20) = 0
<=>               0              +                   5x                      -          22         = 0
<=> 5x = 0 + 22
<=> x = 22/5
Vậy (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4) = 22/5
vô trang mình 5 sao nha

b, (8x - 3)(3x + 2) - (4x - 7)(x + 4) = (2x + 1)(5x - 1)
<=> 24x^2 + 7x - 6 - (4x^2 + 9x - 28) = 10x^2 + 3x - 1
<=> 10x^2 - 5x + 23 = 0
<=> (x - 5/2)^2 + 9x^2 + 23 - 25/4
<=> (x - 5/2)^2 + 9x^2 + 67/4 ≥ 67/4
Dấu "=" xảy ra khi x - 5/2 = 0
<=> x = 5/2
hoặc : 9x^2 = 0
<=> x = 0
Vậy x = 0; x = 5/2
vô trang mình 5 sao nha

b,(8x-3)(3x+2)-(4x-7)(x+4)=(2x+1)(5x-1)
<=> 24x^2 + 16x - 9x - 6 - 4x^2 - 16x + 7x + 28 = 10x^2 - 2x + 5x - 1
<=> 24x^2 - 4x^2 + 16x - 9x - 16x + 7x - 6 + 28 - 10x^2 + 2x - 5x + 1 = 0
<=> 24x^2 - 4x^2 - 10x^2 + 16x - 9x - 16x + 7x + 2x - 5x - 6 + 28 + 1 = 0
<=> 10x^2 - 5x + 23 = 0
Ta có : Δ = b^2 - 4ac = (-5)^2 - 4.10.23 = -895
Do Δ < 0 nên phương trình vô nghiệm

c,2x^2+3(x-1)(x+1) = 0
<=> 2x^2 + 3(x^2 - 1) = 0
<=> 2x^2 + 3x^2 - 3 = 0
<=> 5x^2 - 3 = 0
<=> 5x^2 = 3
<=> x^2 = 3/5
<=> x = ± √15 / 5

a,(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
=>(2x+3)(x-4)+(x-5)(x-2)-(3x-5)(x-4)=0
=>(x-4)(2x+3-3x+5)+(x-5)(x-2)=0
=>(x-4)(-x+8)+(x-5)(x-2)=0
=>-x^2+8x+4x-32+x^2-2x-5x+10=0
=>5x-22=0
=>5x=22
=> x=22/5
Vậy pt có nghiệm x=22/5
b,(8x-3)(3x+2)-(4x-7)(x+4)=(2x+1)(5x-1)
=>24x^2+16x-9x-6-(4x^2+16x-7x-28)=10x^2-2x+5x-1
=>10x^2 -5x+23=0
=>10(x^2-x/2+1/16)+23-5/4=0
<=>10(x-1/4)^2 +87/4=0
Do (x-1/4)^2≥ 0=>10(x-1/4)^2 +87/4>0
=>pt vô nghiệm

c,2x^2+3(x-1)(x+1)=0
=> 2x^2 + 3(x^2 - 1) = 0
=> 2x^2 + 3x^2 - 3 = 0
=> 5x^2 - 3 = 0
=> 5x^2 = 3
=> x^2 = 3/5
=>|x|=√(3/5)
nếu x>0=>x=√(3/5)
Nếu x<0=>x=-√(3/5)
Vậy pt có nghiệm x = ± √(3/5)

các bạn ơi ý c mik vt thiếu
câu của nó là 2x^2 + 3 ( x- 1 ) ( x + 1 ) = 5x ( x + 1 )