Question

1) Giải các pt sau :

a) \(\frac{2x-1}{2}-1=\frac{x^2+x-3}{x-1}-\frac{5x-2}{2-2x}\)

b) 3 (x - 1)\(^2\) - x\(^2\) + 1= 0

c) 4x\(^2\) - 3x - 1= 0

d) 4x\(^2\) - 4x= 3

Answer 1

a)⇔ 2x-1/2 -1 = x²+x-3/x-1 - 5x-2/2.(x-1)
⇔ ( 2x-1 ).(x-1) - 2.(x-1)=2.(x²+x-3) - (5x-2)
⇔2x²-3x+1-2x+2=2x²+2x-6-5x+2
⇔2x²-3x+1-2x+2-2x²-2x+6+5x-2=0
⇔-2x+7=0
⇔x=7/2
Vậy ....
b) ⇔3.(x-1)²-(x-1).(x+1)=0
⇔ (x-1).(3x-3-x-1)=0
⇔ (x-1).(2x-4)=0
⇔x=1 hoặc x=2
Vậy....
c) ⇔ 4x²-4x+x-1=0
⇔4x(x-1)+(x-1)=0
⇔(x-1)(4x+1)=0
⇔x=1 hoặc x=-1/4
Vậy....
d) ⇔4x²-4x-3=0
⇔ 4x²-6x+2x-3 = 0
⇔ 2x( 2x-3)+(2x-3)=0
⇔ (2x+3)(2x+1)=0
⇔ x=-3/2 hoặc x=-1/2
vậy ....

Answer 2

\(a,\frac{2x-1}{2}-1=\frac{x^2+x-3}{x-1}-\frac{5x-2}{2-2x}ĐKXĐ:x\ne1\)

\(\left(2x-1\right)\left(x-1\right)\left(1-x\right)-2\left(x-1\right)\left(1-x\right)=2\left(x^2+x-3\right)\left(1-x\right)-\left(5x-2\right)\left(x-1\right)\)

\(7x^2-8x+3=-5x^2+15x-8\)

\(7x^2-8x+3+5x^2-15x+8=0\)

\(12x^2-23x+11=0\)

\(\left(12x-11\right)\left(x-1\right)=0\)

\(\left[{}\begin{matrix}12x=11\\x=1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\frac{11}{12}\\x=1\end{matrix}\right.\)Theo ĐKXĐ => x= \(\frac{11}{12}\)