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\(\frac{-3}{5}< x< \frac{1}{7}\Leftrightarrow\left\{{}\begin{matrix}x+\frac{3}{5}>0\\x-\frac{1}{7}< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left|x+\frac{3}{5}\right|=x+\frac{3}{5}\\\left|x-\frac{1}{7}\right|=-x+\frac{1}{7}\end{matrix}\right.\)
Khi đó: \(C=x+\frac{3}{5}-x+\frac{1}{7}+\frac{4}{5}=\frac{54}{35}\)
Vậy...
Khi \(B=-\frac{3}{5}\)ta có :
\(B=\left|x-\frac{1}{7}\right|-\left|x+\frac{3}{5}\right|+\frac{4}{5}\)
\(B=\left|-\frac{3}{5}-\frac{1}{7}\right|-\left|-\frac{3}{5}+\frac{3}{5}\right|+\frac{4}{5}\)
\(B=-\frac{26}{35}-0+\frac{4}{5}\)
\(B=-\frac{26}{35}+\frac{4}{5}\)
\(B=\frac{2}{35}\)
\(B=\left|\frac{-3}{5}-\frac{1}{7}\right|-\left|\frac{-3}{5}+\frac{3}{5}\right|+\frac{4}{5}=\frac{26}{35}+\frac{4}{5}=\frac{2}{35}\)
a: x>-3/5 nên x+3/5>0
x<1/7 nên x-1/7<0
A=1/7-x-(x+3/5)+4/5
=1/7-2x-3/5+4/5
=-2x+12/35
b: \(B=\left|-x+\dfrac{1}{7}\right|+\left|-x-\dfrac{3}{5}\right|-\dfrac{2}{6}\)
\(=\left|x-\dfrac{1}{7}\right|+\left|x+\dfrac{3}{5}\right|-\dfrac{1}{3}\)
-3/5<x<1/7
nên x-1/7<0; x+3/5>0
\(B=\dfrac{1}{7}-x+x+\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{43}{105}\)
c: \(C=\left|\dfrac{11}{5}-x\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
\(=\left|x-\dfrac{11}{5}\right|+\left|x-\dfrac{1}{5}\right|+\dfrac{41}{5}\)
Nếu 1/5<x<11/5 nên x-1/5>0; x-11/5<0
\(C=\dfrac{11}{5}-x+x-\dfrac{1}{5}+\dfrac{41}{5}=\dfrac{51}{5}\)