Cho ∆ABC vuông tại A , tia phân giác của góc B cắt AC tại M . Trên cạnh BC lấy D sao cho AD = AB a, chứng minh ∆ABM=∆DBM b, chứng MD,
cTia ba cắt de tại e cmr ad song song với ce vuông góc với BC Cần gấp ạHãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = \(\dfrac{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}\)
Xét TS = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) \(\dfrac{2020}{3}\) +... + \(\dfrac{1}{2022}\)
TS = (1 + \(\dfrac{2021}{2}\)) + (1 + \(\dfrac{2020}{3}\)) + ... + ( 1 + \(\dfrac{1}{2022}\)) + 1
TS = \(\dfrac{2023}{2}\) + \(\dfrac{2023}{3}\) +...+ \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2023}\)
TS = 2023.(\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{2023}\))
A = \(\dfrac{2023.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}\)
A = 2023
Lời giải:
a. Thể tích cái bánh: $6\times 8\times 3:2=72$ (cm3)
b.
Độ dài cạnh chéo miếng bánh: $\sqrt{6^2+8^2}=10$ (cm)
Diện tích vật liệu cần dùng chính là diện tích toàn phần của cái bánh và bằng:
$6.8+6.3+3.8+10.3=120$ (cm2)
Viết đoạn văn nêu cảm nghĩ về một câu ca dao hoặc tục ngữ của Hải Dương mà em thích.
Giúp mik với ah
`#3107.101107`
Ta có: `\text{AB // CD}`
\(\Rightarrow\) \(\widehat{\text{BAC}}=\widehat{\text{DCA}}\)
Vì `\text{AD // BC}`
\(\Rightarrow\widehat{\text{DAC}}=\widehat{\text{BCA}}\)
Xét `\Delta ABC` và `\Delta CDA` :
\(\widehat{\text{BAC}}=\widehat{\text{DCA}}\)
\(\text{AC chung}\)
\(\widehat{\text{DAC}}=\widehat{\text{BCA}}\)
\(\Rightarrow\Delta\text{ABC = }\Delta\text{CDA (g - c - g).}\)
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Hope this helps you.
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