\(A=5002.5002\)

\(A=5002.\left(5000+2\right)\)

\(A=5002.5000+5002.2\)

\(B=5000.5004\)

\(B=5000.\left(5002+2\right)\)

\(B=5000.5002+5000.2\)

VẬY B < A

1+1=2

tk nhé

a)Gọi \(\widehat{CBx}\) và \(\widehat{BCy}\) lần lượt là 2 góc ngoài đỉnh B và C

Ta có:

\(\widehat{ABC}+\widehat{ACB}=180^0-\widehat{BAC}=180^0-50^0=130^0\)

\(\Rightarrow\frac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=\frac{1}{2}.130^0=65^0\)

\(\Rightarrow180^0-\frac{1}{2}\left(\widehat{ABC}+\widehat{ACB}\right)=180^0-65^0=115^0\)

\(\Rightarrow`180^0-\left(\widehat{IBC} +\widehat{ICB}\right)=115^0\)

\(\Rightarrow\widehat{BIC}=115^0\)

\(\Rightarrow\widehat{BKC}=180^0-\frac{1}{2}\left(360^0-130^0\right)=180^0-\frac{1}{2}.230=180^0-115=65^0\)b) Ta có:

\(\widehat{BDK}=180^0-\left(\widehat{DBK}+\widehat{DKB}\right)=180^0-\left(90^0+65^0\right)=180^0-155^0=25^0\)Hay \(\widehat{BDC}=25^0\)

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