lỗi r nha bn

lỗi

a: Xét ΔAMC và ΔBMD có 

MA=MB

\(\widehat{AMC}=\widehat{BMD}\)

MC=MD

Do đó: ΔAMC=ΔBMD

\(\dfrac{3^7.8^3}{6^5.12^2}=\dfrac{3^7.\left(2^3\right)^3}{\left(2.3\right)^5.\left(2^2.3\right)^2}=\dfrac{3^7.2^9}{2^5.3^5.2^4.3^2}=\dfrac{3^7.2^9}{3^7.2^9}=1\)

\(\dfrac{15^3+3.15^2+5^2.3^3}{7.5.3^3}=\dfrac{3^3.5^3+3.3^2.5^2+3^3.5^2}{7.5.3^3}=\dfrac{5.3^3\left(5^2+5+5\right)}{7.5.3^3}=\dfrac{25+5+5}{7}=5\)

c) \(\dfrac{3^7\cdot8^3}{6^5\cdot12^2}=\dfrac{3^7\cdot2^9}{2^5\cdot3^5\cdot3^2\cdot2^4}=1\)

d) \(\dfrac{15^3+3\cdot15^2+5^2\cdot3^3}{7\cdot5\cdot3^3}=\dfrac{15^3+3\cdot15^2+15^2\cdot3}{7\cdot5\cdot3^3}\)

\(=\dfrac{15^2\cdot\left(15+6\right)}{7\cdot5\cdot3^3}=\dfrac{3^3\cdot5^2\cdot7}{7\cdot5\cdot3^3}=5\)

từ điểm B kẻ \(Bz//Cy=>\angle\left(BCy\right)+\angle\left(CBz\right)=180^o\)(góc trong cùng phía)

\(=>\angle\left(CBz\right)=180^o-130^o=50^o\)

\(=>\angle\left(ABz\right)=\angle\left(ABC\right)+\angle\left(CBz\right)=50^o+72^o=122^o\)

\(=>\angle\left(BAx\right)+\angle\left(ABz\right)=180^o\)

mà 2 góc này ở vị trí trong cùng phía

\(=>Ax//Bz=>Ax//Cy\)

kẻ Bz//Ax=>Bz//Cy

\(=>\angle\left(BAx\right)+\angle\left(ABz\right)=180^o\)(tgosc trong cùng phía)

\(=>\angle\left(ABz\right)=180^o-100^o=80^o\)

\(=>\angle\left(CBz\right)=80+40=120^o=\angle\left(BCy\right)\)(so le trong)