Question

giups mk bài toán nha

Answer 1

Bài 3.2:

a: ta có: |x+9|=2x

\(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\\left(2x\right)^2-\left(x+9\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(2x-x-9\right)\left(2x+x+9\right)=0\end{matrix}\right.\)

=>x=9

b: ta có: \(\left|5x\right|=3x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(5x-3x+2\right)\left(5x+3x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(2x+2\right)\left(8x-2\right)=0\end{matrix}\right.\)

hay \(x\in\varnothing\)

c: Ta có: \(\Leftrightarrow\left|x+6\right|=2x+9\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{9}{2}\\\left(2x+9-x-6\right)\left(2x+9+x+6\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{2}\\\left(x+3\right)\left(3x+15\right)=0\end{matrix}\right.\Leftrightarrow x=-3\)

d: \(\Leftrightarrow\left|2x-3\right|=21-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =21\\\left(21-x-2x+3\right)\left(21-x+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =21\\\left(24-3x\right)\left(x+18\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{8;-18\right\}\)