Question

a,(2x+1)² = (x-1)²

b,(x²-5)(x+3)=0

c, 3(x-1)(2x-1)=5(x+8)(x-1)

d, x³ + x² +x+1=0

Answer 1

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

=>2x+1=x-1 hoặc 2x+1=1-x

=>x=-2 hoặc x=0

b: \(\left(x^2-5\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow x\in\left\{\sqrt{5};-\sqrt{5};-3\right\}\)

c: \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)

hay \(x\in\left\{1;43\right\}\)

d: \(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

=>x+1=0

hay x=-1