Trình bày rõ ràng hộ em ạ
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a) 973
b) 1011101
c) 58
e) 19039
f) 11111101
g) 43
k) 1EA9
l) 401
m) 111001101011
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Câu 5:
a: Xét tứ giác AHMK có
\(\widehat{AHM}=\widehat{AKM}=\widehat{KAH}=90^0\)
Do đó: AHMK là hình chữ nhật
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Ta có:
\(A=\dfrac{\sqrt{x}-x}{\sqrt{x}-1}=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)
Vậy \(A=-\sqrt{x}\)
\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)
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\(\dfrac{2}{7}\cdot\dfrac{14}{15}\cdot\dfrac{5}{6}:\dfrac{3}{8}:\dfrac{2020}{2021}\)
\(=\dfrac{2}{7}\cdot\dfrac{14}{15}\cdot\dfrac{5}{6}\cdot\dfrac{8}{3}\cdot\dfrac{2021}{2020}\)
\(=\dfrac{14}{7}\cdot\dfrac{5}{15}\cdot\dfrac{8}{6}\cdot\dfrac{2}{3}\cdot\dfrac{2021}{2020}\)
\(=2\cdot\dfrac{1}{3}\cdot\dfrac{4}{3}\cdot\dfrac{2}{3}\cdot\dfrac{2021}{2020}\)
\(=\dfrac{8084}{13635}\)
\(\widehat{E}=53^0\)
Áp dụng tslg trong tam giác DEF vuông tại D:
\(tanE=\dfrac{DF}{ED}=\dfrac{4}{3}\Rightarrow\widehat{E}\approx53^0\)