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a) Ta có: \(\left|x+7\right|-\left(-8\right)=-25+73\)
\(\Leftrightarrow\left|x+7\right|+8=48\)
\(\Leftrightarrow\left|x+7\right|=40\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=40\\x+7=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=33\\x=-47\end{matrix}\right.\)
Vậy: \(x\in\left\{33;-47\right\}\)
c) Ta có: \(-\left(a-b\right)+\left(b-c\right)-\left(a-c\right)=2b-2a\)
\(\Leftrightarrow-a+b+b-c-a+c=2b-2a\)
\(\Leftrightarrow-2a+2b-2b+2a=0\)
\(\Leftrightarrow0a+0b=0\)(luôn đúng)
Vậy: \(\left\{{}\begin{matrix}a\in Z\\b\in Z\end{matrix}\right.\)
d) Ta có: \(-\left(-a+b+c\right)+\left(b+c-1\right)=-\left(b-a\right)-\left(1-b\right)\)
\(\Leftrightarrow a-b-c+b+c-1=-b+a-1+b\)
\(\Leftrightarrow a-1=a-1\)(luôn đúng)
Vậy: \(\left\{{}\begin{matrix}a\in Z\\b\in Z\\c\in Z\end{matrix}\right.\)
e) Ta có: \(-\left(-a+b+c\right)+\left(b-c+6\right)=a+6\)
\(\Leftrightarrow a-b-c+b-c+6=a+6\)
\(\Leftrightarrow a+6-2c-a-6=0\)
\(\Leftrightarrow-2c=0\)
hay c=0
Vậy: \(\left\{{}\begin{matrix}a\in Z\\b\in Z\\c=0\end{matrix}\right.\)
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A)a +a=a^2 S
B)a-a=0 Đ
C)a.a=2a S
D)a:a=1 Đ