Question

Tìm x , biết

a, 2x² - 4x =0

b, x³ + 2x² + x = 0

c, x²(x+1) + 2x( x+1) = 0

d, x(2x - 3) - 2( 3 - 2x) = 0

áp dụng các phương pháp phân tích đa thức thành nhân tử

Answer 1

a, \(2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

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

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b, \(x^3+2x^2+x=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

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c, \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)

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d, \(x\left(2x-3\right)-2\left(3-2x\right)=0\)

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

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

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

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Answer 2

a) \(2x^2-4x=0\\ \Leftrightarrow2x\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

b) \(x^3+2x^2+x=0\\\Leftrightarrow x\left(x^2+2x+1\right)=0\\ \Leftrightarrow x\left(x+1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\\ \Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

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

Answer 3

\(a,2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

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

\(b,x^3+2x^2+x=0\)

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

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

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

\(c,x^2\left(x+1\right)+2x\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)

\(d,x\left(2x-3\right)-2\left(3-2x\right)=0\)

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

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

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