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\(2x^2-8x+3x-12+x^2-2x-5x+10-3x^2+12x+5x-20\)=0
(\(2x^2+x^2-3x^2\))+(-8x+3x-2x-5x+12x+5x)+(-12+10-20) =0 5x-22 =0
5x = 22
x = \(\dfrac{22}{5}\)
Vậy x= \(\dfrac{22}{5}\)
b) (8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)
(8x-3)(3x+2)-(4x+7)(x+4)-(2x+1)(5x-1)=0
\(24x^2\) +16x-9x-6\(-4x^2\) -16x-7x-28\(-10x^2\) +2x-5x+1=0
(24\(x^2-4x^2-10x^2\))+(16x-9x-16x-7x+2x-5x)+(-6-28+1)=0
10\(x^2-19x-33\)=0
10\(x^2+11x-30x-33=0\)
x(10x+11)-3(10x+11)=0
(x-3) (10x+11)=0
=>x-3=0 => x=3 =>x=3
10x+11=0 10x=-11 x=\(\dfrac{-11}{10}\)
Vậy x=3 hoặc x=\(\dfrac{-11}{10}\)
Tìm x, biết:
a) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
<=> \(2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
<=> \(2x^2-8x+3x+x^2-2x-5x-3x^2+12x+5x=12-10+20\)
<=> \(5x=22\)
<=> \(x=\dfrac{22}{5}\)
Vậy \(S=\left\{\dfrac{22}{5}\right\}\)
bạn ơi xem lại câu b có đúng ko nhá mik thấy nó sao sao í