![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(=>Ax//Bz=>Ax//Cy\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng t/c dãy tỉ số bằng nhau
\(\dfrac{a}{2013}=\dfrac{b}{2012}=\dfrac{c}{2011}=\dfrac{a-c}{2}=\dfrac{a-b}{1}=\dfrac{b-c}{1}\\ \Rightarrow a-c=2\left(a-b\right)=2\left(b-c\right)\)
\(\Rightarrow H=\dfrac{\left[2\left(a-b\right)\right]^4}{\left(a-b\right)^2\left(a-b\right)^2}=\dfrac{16\left(a-b\right)^4}{\left(a-b\right)^4}=16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\c+a=2b\\a+b=2c\end{matrix}\right.\)
\(\Rightarrow P=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=\dfrac{2a}{a}+\dfrac{2b}{b}+\dfrac{2c}{c}=2+2+2=6\)
P=
\(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}=\dfrac{a}{b+c}.\left(\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\right):\left(\dfrac{a}{b+c}\right)=\left(\dfrac{b+c}{a}.\dfrac{a}{b+c}+\dfrac{c+a}{b}.\dfrac{a}{b+c}+\dfrac{a+b}{c}.\dfrac{a}{b+c}\right):\dfrac{a}{b+c}=\left(\dfrac{b+c}{a}.\dfrac{a}{b+c}+\dfrac{c+a}{b}.\dfrac{b}{c+a}+\dfrac{a+b}{c}.\dfrac{c}{a+b}\right):\dfrac{a}{b+c}=\left(1+1+1\right):\dfrac{a}{b+c}=3.\dfrac{b+c}{a}=\dfrac{3b+3c}{a}\)
\(\dfrac{3^7.8^3}{6^5.12^2}=\dfrac{3^7.\left(2^3\right)^3}{\left(2.3\right)^5.\left(2^2.3\right)^2}=\dfrac{3^7.2^9}{2^5.3^5.2^4.3^2}=\dfrac{3^7.2^9}{3^7.2^9}=1\)
\(\dfrac{15^3+3.15^2+5^2.3^3}{7.5.3^3}=\dfrac{3^3.5^3+3.3^2.5^2+3^3.5^2}{7.5.3^3}=\dfrac{5.3^3\left(5^2+5+5\right)}{7.5.3^3}=\dfrac{25+5+5}{7}=5\)
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) \(\dfrac{15^3+3\cdot15^2+5^2\cdot3^3}{7\cdot5\cdot3^3}=\dfrac{15^3+3\cdot15^2+15^2\cdot3}{7\cdot5\cdot3^3}\)
\(=\dfrac{15^2\cdot\left(15+6\right)}{7\cdot5\cdot3^3}=\dfrac{3^3\cdot5^2\cdot7}{7\cdot5\cdot3^3}=5\)