Question

Tìm x :

a)\(x\left(8x-2\right)-8x^2+12=0\) 

b) \(x\left(4x-5\right)-\left(2x+1\right)^2=0\)

 

Answer 1

\(a,x\left(8x-2\right)-8x^2+12=0\)

\(\Rightarrow8x^2-2x-8x^2+12=0\)

\(\Rightarrow-2x+12=0\)

\(\Rightarrow-2x=-12\)

\(\Rightarrow x=6\)

\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Rightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Rightarrow-9x-1=0\)

\(\Rightarrow-9x=1\)

\(\Rightarrow x=\frac{-1}{9}\)

Answer 2

a) x(8 - 2) - 8x² + 12 = 0

x(8 - 2) - 8x² = 12 - 0

x(8 - 2) - 8x² = 12

2x = 12

x = 6

b) x(4x - 5) - (2x + 1)² = 0

9x - 1 = 0

9x = 0 + 1

9x = 1

x = -1/9

Answer 3

\(a,x\left(8x-2\right)-8x^2+12=0\)

\(\Leftrightarrow8x^2-2x-8x^2+12=0\)

\(\Leftrightarrow-2x+12=0\)

\(\Leftrightarrow-2x=-12\)

\(\Leftrightarrow x=6\)

\(b,x\left(4x-5\right)-\left(2x+1\right)^2=0\)

\(\Leftrightarrow4x^2-5x-4x^2-4x-1=0\)

\(\Leftrightarrow-9x-1=0\)

\(\Leftrightarrow-9x=1\)

\(\Leftrightarrow x=-\frac{1}{9}\)