Question

tìm nghiệm

a,x^2-2x

b,x+3(x-1)+8(3-x)

c,x^-5x+42

Answer 1

quế bênh phải là bao nhiêu

Answer 2

\(a,x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\Rightarrow x=2\end{matrix}\right.\)

b, \(x+3\left(x-1\right)+8\left(3-x\right)=0\)

\(\Leftrightarrow x+3x-3+24-8x=0\)

\(\Leftrightarrow-4x=21\Leftrightarrow x=-\dfrac{21}{4}\)

c, \(x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)