Question

[huy hongnm]

Tìm x

a, \(\frac{1}{3}x-\frac{2}{5}.\left(x+1\right)=0\)

b, \(\frac{x+2}{0,5}=\frac{2x+1}{2}\)

c, \(|x+\frac{1}{5}|-4=2\)

d, \(\left(\frac{1}{4}x-1,5\right)+\left(\frac{5}{6}x-3\right)-\left(\frac{5}{8}x-0,5\right)=-4,5\)

Answer 1

[tthnew]

a) Quy đồng lên đi.

b) \(\frac{x+2}{0.5}=\frac{2x+1}{2}\Leftrightarrow\frac{x+2}{\left(\frac{1}{2}\right)}=\frac{2x+1}{2}\)

\(\Leftrightarrow2x+4=\frac{2x+1}{2}\Leftrightarrow4x+8=2x+1\)

\(\Leftrightarrow x=-\frac{7}{2}\)

c) \(\Leftrightarrow\left|x+\frac{1}{5}\right|=6\). VỚi x >= -1/5 thì:

\(x+\frac{1}{5}=6\Leftrightarrow x=\frac{29}{5}\left(TM\right)\)

Với x < -1/5 thì \(-x-\frac{1}{5}=6\Leftrightarrow x=-\frac{31}{5}\left(TM\right)\)

d) TƯơng tự ý a, quy đồng lên thôi (mẫu chung là 24 thì phải)

Answer 2

[Vũ Minh Tuấn]

c) \(\left|x+\frac{1}{5}\right|-4=2\)

=> \(\left|x+\frac{1}{5}\right|=2+4\)

=> \(\left|x+\frac{1}{5}\right|=6\)

=> \(\left\{{}\begin{matrix}x+\frac{1}{5}=6\\x+\frac{1}{5}=-6\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=6-\frac{1}{5}\\x=\left(-6\right)-\frac{1}{5}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{29}{5}\\x=-\frac{31}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{29}{5};-\frac{31}{5}\right\}\).

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!