Giải giúp bài này ạ
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Answer:
\(23\frac{1}{3}:\left(-\frac{5}{7}\right)-13\frac{1}{3}:\left(-\frac{5}{7}\right)\)
\(=23\frac{1}{3}.\left(-\frac{7}{5}\right)-13\frac{1}{3}.\left(-\frac{7}{5}\right)\)
\(=\left(-\frac{7}{5}\right).\left(23\frac{1}{3}-13\frac{1}{3}\right)\)
\(=\left(-\frac{7}{5}\right).10\)
\(=-14\)
\(25.\left(-\frac{1}{5}\right)+\frac{1}{2}-2.\left(-\frac{1}{2}\right)^2-\sqrt{\frac{1}{4}}\)
\(=-5+\frac{1}{5}-2.\frac{1}{4}-\frac{1}{2}\)
\(=-\frac{25}{4}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)
\(=-\frac{24}{25}-1\)
\(=\frac{-29}{5}\)
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Answer:
A C B 50 độ
Ta xét tam giác ABC
AB = AC (gt)
=> Tam giác ABC cân tại A
\(\Rightarrow\widehat{ABC}=\widehat{ACB}\)
Ta có: \(\widehat{BAC}+\widehat{ABC}+\widehat{BCA}=180^o\)
Mà \(\widehat{ABC}=\widehat{ACB}\) (chứng minh trên)
\(\Rightarrow\widehat{BAC}+2\widehat{ABC}=180^o\)
\(\Rightarrow2\widehat{ABC}=180^o-\widehat{BAC}\)
\(\Rightarrow\widehat{ABC}=\widehat{ACB}=\frac{180^o-50^o}{2}=\frac{130^o}{2}=65^o\)