Question

Giải các phương trình:

a) |x + 5| = 3x + 1;

b) |-5x| = 2x + 21.

Answer 1

bạn tự kl nhé 

a, \(\left|x+5\right|=3x+1\)

TH1 : \(x+5=3x+1\Leftrightarrow-2x=-4\Leftrightarrow x=2\)

TH2 : \(x+5=-3x-1\Leftrightarrow4x=-6\Leftrightarrow x=-\dfrac{3}{2}\)( ktm )

b, \(\left|-5x\right|=2x+21\)

TH1 : \(5x=2x+21\Leftrightarrow3x=21\Leftrightarrow x=7\)

TH2 : \(5x=-2x-21\Leftrightarrow7x=-21\Leftrightarrow x=-3\)

Answer 2

khó nhìn quá ko

 

Answer 3

a) Ta có: |x+5|=3x+1

\(\Leftrightarrow\left[{}\begin{matrix}x+5=3x+1\left(x\ge-5\right)\\x+5=-3x-1\left(x< -5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3x=1-5\\x+3x=-1-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-2x=-4\\4x=-6\end{matrix}\right.\)

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

Vậy: S={2}

Answer 4

b) Ta có: \(\left|-5x\right|=2x+21\)

\(\Leftrightarrow\left[{}\begin{matrix}-5x=2x+1\left(x\le0\right)\\5x=2x+1\left(x>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x-2x=1\\5x-2x=1\end{matrix}\right.\)

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

Vậy: \(S=\left\{-\dfrac{1}{7};\dfrac{1}{3}\right\}\)