47:AD//BCAD⊥ ABDo đó: BC⊥BA=>\(\hat{ABC}=90^0\) AD//BC=>\(\hat{ADC}+\hat{BCD}=180^0\) (hai góc trong cùng phía)=>\(\hat{ADC}=180^0-130^0=50^0\) BÀi 46:a: \(\frac{x}{27}=-\frac{2}{3,6}\) =>\(\frac{x}{27}=-\frac59\) =>\(x=-27\cdot\frac59=-15\) b: \(-0,52:x=-9,36:16,38\) =>\(\frac{0.52}{x}=\frac{936}{1638}=\frac47\) =>\(x=7\cdot\frac{0.52}{4}=7\cdot0,13=0,91\) c: \(\frac{4\frac14}{2\frac78}=\frac{x}{1,61}\) =>\(\frac{x}{1,61}=\frac{17}{4}:\frac{23}{8}=\frac{17}{4}\cdot\frac{8}{23}=\frac{34}{23}\) =>\(x=1,61\cdot\frac{34}{23}=2,38\)Câu trả lời: 47:AD//BCAD⊥ ABDo đó: BC⊥BA=>\(\hat{ABC}=90^0\) AD//BC=>\(\hat{ADC}+\hat{BCD}=180^0\) (hai góc trong cùng phía)=>\(\hat{ADC}=180^0-130^0=50^0\) BÀi 46:a: \(\frac{x}{27}=-\frac{2}{3,6}\) =>\(\frac{x}{27}=-\frac59\) =>\(x=-27\cdot\frac59=-15\) b: \(-0,52:x=-9,36:16,38\) =>\(\frac{0.52}{x}=\frac{936}{1638}=\frac47\) =>\(x=7\cdot\frac{0.52}{4}=7\cdot0,13=0,91\) c: \(\frac{4\frac14}{2\frac78}=\frac{x}{1,61}\) =>\(\frac{x}{1,61}=\frac{17}{4}:\frac{23}{8}=\frac{17}{4}\cdot\frac{8}{23}=\frac{34}{23}\) =>\(x=1,61\cdot\frac{34}{23}=2,38\)