Câu a nếu bn chứng minh đc a // c r thì bn sẽ lm là:

Vì \(\widehat{IGB}.và.\widehat{GBC}.\text{bù nhau}.\text{nên c//b}\)

a // c; c//b => a // b

\(\text{Xin cái tick}\)

IGB và GBC là 2 góc trong cùng phía bù nhau nhá mk quên

\(8,\\ b,4^{333}=\left(4^3\right)^{111}=64^{111}< 81^{111}=\left(3^4\right)^{111}=3^{444}\\ c,2^{500}=\left(2^5\right)^{100}=32^{100}>25^{100}=\left(5^2\right)^{100}=5^{200}\\ d,2^{375}=\left(2^3\right)^{125}=8^{125}< 9^{125}=\left(3^2\right)^{125}=3^{250}\)

b: \(4^{333}=\left(4^3\right)^{111}=64^{111}\)

\(3^{444}=\left(3^4\right)^{111}=81^{111}\)

mà 64<81

nên \(4^{333}< 3^{444}\)

c: \(2^{500}=\left(2^5\right)^{100}=32^{100}\)

\(5^{200}=\left(5^2\right)^{100}=25^{100}\)

mà 32>25

nên \(2^{500}>5^{200}\)

B=\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{19}{20}\)

\(=\dfrac{1.2.3....19}{2.3.4.....20}\)

\(=\dfrac{1.2.3....19:\left(2.3.....19\right)}{2.3.4.....20:\left(2.3.4.....19\right)}\)

\(=\dfrac{1}{20}\)

Câu 2:
1.Do xy\(\perp\)CD mà zt\(\perp\)CD
\(=>xy//zt\)
2.\(=>\widehat{ABt}=\widehat{mAy}=46\left(haigocdongvi\right)\)