Bài 4: Phương trình tích

HP

a) (2x-1)^2+(2-x)(2x-1)=0

b) x^3+1=x(x+1)

c) x^2-5x+6=0

ND
20 tháng 2 2018 lúc 17:23

a.

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

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

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

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

b.

\(x^3+1=x\left(x+1\right)\)

\(\Leftrightarrow x^3+1=x^2+x\)

\(\Leftrightarrow x^3+1-x^2-x=0\)

\(\Leftrightarrow x^3+x^2-2x^2-2x+x+1=0\)

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

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

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

\(\Leftrightarrow x=-1\) hoặc \(x=1\)

c.

\(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

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

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

Bình luận (0)
GN
20 tháng 2 2018 lúc 17:25

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

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

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

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

Vậy ...............

b) \(x^3+1=x\left(x+1\right)\)

\(\Leftrightarrow x^3+1=x^2+x\)

\(\Leftrightarrow x^3-x^2-x+1=0\)

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

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

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

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

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

Vậy ..............

c) \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

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

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

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

Vậy ......................

Bình luận (2)

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