Bài 1:Tính
a,-\(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(-\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
b,\(12\frac{3}{5}:\left(-\frac{5}{7}\right)+2\frac{2}{5}:\left(-\frac{5}{7}\right)\)
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\(\left|5x-3\right|=\left|7-x\right|\)
\(\Rightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x+x=7+3\\5x-x=\left(-7\right)+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
`|5x-3|=|7-x|`
TH1 : `5x-3=7-x -> 5x+x=7+3-> 6x=10 ->x=5/3`
TH2 : `5x-3=-7 +x -> 5x-x=-7+3 -> 4x=-4 ->x=-1`
Vậy `x=5/3,x=-1`
Vì góc CBA' kề bù với góc ABC.
\(\Rightarrow\)CBA' + ABC = 180o
Thay ABC = 56o
\(\Rightarrow\)CBA' + 56o = 180o
\(\Rightarrow\)CBA' = 180o - 56o
\(\Rightarrow\)CBA' = 124o
a) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(-\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
\(=-\frac{13}{25}+\frac{4}{191}-\frac{2}{51}-\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\)
\(=\left(-\frac{13}{25}+\frac{3}{5}\right)+\left(\frac{4}{191}-\frac{4}{191}\right)-\left(\frac{2}{51}+\frac{2}{51}\right)\)
\(=\frac{2}{25}+0-0=\frac{2}{25}\)
b) \(12\frac{3}{5}:\left(-\frac{5}{7}\right)+2\frac{2}{5}:\left(-\frac{5}{7}\right)\)
\(=\frac{63}{5}.\frac{-7}{5}+\frac{12}{5}.\frac{-7}{5}\)
\(=\left(\frac{63}{5}+\frac{12}{5}\right).\frac{-7}{5}\)
\(=15.\frac{-7}{5}=-21\)