Question

tìm x:\(\left(x-2\right)^2-\left(x+1\right)\left(x-1\right)=0\)

Answer 1

\(\left(x-2\right)^2-\left(x+1\right)\left(x-1\right)=0\\ =>x^2-4x+4-x^2+1=0\\ =>-4x+5=0\\ =>-4x=-5\\ =>x=\dfrac{5}{4}\)

Answer 2

`(x-2)^2 -(x+1)(x-1)=0`

`<=> (x^2-4x+4)-(x^2-x+x-1)=0`

`<=> x^2-4x+4-x^2+x-x+1=0`

`<=> -4x +5=0`

`<=>-4x=-5`

`<=>x=5/4`

Answer 3

=>x^2-4x+4-x^2+1=0

=>-4x+5=0

=>-4x=-5

=>x=5/4