Question

Tìm x:10x^2-x(x+2)+8x+1 = 0

Answer 1

giúp mình với ạ

 

Answer 2

\(10x^2-x\left(x+2\right)+8x+1=0\\ \Rightarrow10x^2-x^2-2x+8x+1=0\\ \Rightarrow9x^2+6x+1=0\\ \Rightarrow\left(3x+1\right)^2=0\\ \Rightarrow3x+1=0\\ \Rightarrow x=-\dfrac{1}{3}\)

Answer 3

Ta có: \(10x^2-x\left(x+2\right)+8x+1=0\)

\(\Leftrightarrow9x^2+6x+1=0\)

\(\Leftrightarrow3x+1=0\)

hay \(x=-\dfrac{1}{3}\)