Question

[vung nguyen thi]

Giải các phương trình sau

a/ \(\sqrt{x^2+6x+9}\) = \(\left|2x-1\right|\)

b/ \(\left|x^2-2x\right|\)= \(\left|2x^2-x-2\right|\)

c/ \(\left|3x^2-2x\right|\)= \(\left|6-x^2\right|\)

d/ \(\left|x^2-4x-5\right|\) = \(\left|2x^2-3x-5\right|\)

e/ \(\left|5x+1\right|\) = \(\left|2x-3\right|\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: =>|x+3|=|2x-1|

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+3\\2x-1=-x-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=-2\end{matrix}\right.\Leftrightarrow x\in\left\{4;-\dfrac{2}{3}\right\}\)

b: \(\left|x^2-2x\right|=\left|2x^2-x-2\right|\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-1\right)=0\\\left(x+1\right)\left(3x-2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1;-1;\dfrac{2}{3}\right\}\)

c: \(\left|3x^2-2x\right|=\left|6-x^2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=6-x^2\\3x^2-2x=x^2-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x^2-2x-6=0\\2x^2-2x+6=0\end{matrix}\right.\)

\(\Leftrightarrow2x^2-x-3=0\)

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

=>x=3/2 hoặc x=-1

d: \(\left|2x^2-3x-5\right|=\left|x^2-4x-5\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=x^2-4x-5\\2x^2-3x-5=4x+5-x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+x=0\\3x^2-7x-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\3x^2-10x+3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\left(x+1\right)=0\\\left(3x-10\right)\left(x+1\right)=0\end{matrix}\right.\)

hay \(x\in\left\{\dfrac{10}{3};-1\right\}\)

e: |5x+1|=|2x-3|

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=2x-3\\5x+1=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-4\\7x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{2}{7}\end{matrix}\right.\)