Question

tìm x

a) \(x.\left(x-3\right)-\left(x+2\right)^2=3\)

b) \(\left(2x-1\right)^2-\left(4x^2-2\right)=\frac{1}{2}\)

GIÚP MÌNH NHA

Answer 1

\(a,x\left(x-3\right)-\left(x+2\right)^2=3=>x^2-3x-\left(x^2+4x+4\right)=3\)

\(=>x^2-3x-x^2-4x-4=3=>-7x-4=3=>-7x=7=>x=-1\)

\(b,\left(2x-1\right)^2-\left(4x^2-2\right)=\frac{1}{2}=>4x^2-4x+1-4x^2+2=\frac{1}{2}\)

\(=>-4x+3=\frac{1}{2}=>-4x=\frac{1}{2}-3=-\frac{5}{2}=>x=\frac{5}{8}\)

Answer 2

a) x.( x - 3 ) - ( x + 2 ) ²  = 3

   x²  - 3x - x² - 4x - 4 = 3

   - 7x - 4 = 3

   - 7x = 3 + 4

   - 7x = 7

   x = 7 : ( - 7 )

   x = - 1

   Vậy x = 1. 

b) ( 2x - 1 )² - ( 4x² - 2 ) = \(\frac{1}{2}\)

    4x² - 4x + 1 - 4x² + 2 = \(\frac{1}{2}\)

    - 4x + 3 = \(\frac{1}{2}\)

    - 4x = \(\frac{1}{2}\) \(-3\)

    - 4x = \(\frac{-5}{2}\)

    x = \(\frac{-5}{2}:4\)

    x = \(\frac{-5}{8}\)

Vậy x = \(\frac{-5}{8}\)