Question

[Nguyễn Minh Chiến]

Giups tớ với

Answer 1

[Nguyễn Việt Lâm]

a.

\(x\left(x-1\right)+x-1=0\)

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

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

b.

\(5x\left(x-7\right)+3x-21=0\)

\(\Leftrightarrow5x\left(x-7\right)+3\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(5x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-7=0\\5x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

Answer 2

[Toru]

a) \(x\left(x-1\right)+x-1=0\)

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

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

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

\(---\)

b) \(5x\left(x-7\right)+3x-21=0\)

\(\Leftrightarrow5x\left(x-7\right)+\left(3x-21\right)=0\)

\(\Leftrightarrow5x\left(x-7\right)+3\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{3}{5}\end{matrix}\right.\)

\(---\)

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

\(\Leftrightarrow\left(5x^2-5\right)-4\left(x^2-2\cdot x\cdot1+1^2\right)=0\)

\(\Leftrightarrow5\left(x^2-1^2\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+5-4x+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\)

\(---\)

d) \(4x^2-9-\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(4x^2-9\right)-\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[\left(2x\right)^2-3^2\right]-\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)-\left(2x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left[\left(2x+3\right)-\left(x-4\right)\right]=0\)

\(\Leftrightarrow\left(2x-3\right)\left(2x+3-x+4\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-7\end{matrix}\right.\)

\(\text{#}Toru\)