Mn ơi giúp mình bài này với ạ, mik cảm ơn ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(\dfrac{1}{49}-\dfrac{1}{9}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)=0\)
Bạn sai cta nhé, sắp not xắp
Tham khảo
1. A special kind of tea is sold here
2. All the cars and trucks have been searched
3. He was put in prison by the goverment last years
4. We will be met by her parents at the station tomorrow
5. A meeting is being held by Mr. Brown in the hall
Bài 2
1 Tea can't be made with cold water
2 Some of my books have been taken away.
3 Some pictures were taken away by the boys.
4 This room may be used for the classroom.
5 This machine mustn't be used after 5:30 p.m
6 Mr Cole used to be visited at weekends by John.
7 All the homework ought to be done by her
Do \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Rightarrow1-\dfrac{b}{a}=1-\dfrac{d}{c}\Rightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\) (đpcm)
cặp : Ea// Fb (vì góc e +góc f =180 mà 2 góc này ở vị trí trong cùng phía)
cặp Fb // DC (vì có góc F = góc D (=110) mà 2 góc này ở vị trí đồng vị)
cặp : Ea //DC vì Ea // Fb, Fb //DC (tính chất bắc cầu)
\(\\ \)
c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\)
Ở nơi x=9/4-1/2 là x-9/4-1/2 nha
a. -1,5 + 2x = 2,5
<=> 2x = 2,5 + 1,5
<=> 2x = 4
<=> x = 2
b. \(\dfrac{3}{2}\left(x+5\right)-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{3}{2}x+\dfrac{15}{2}-\dfrac{1}{2}=\dfrac{4}{3}\)
<=> \(\dfrac{9x}{6}+\dfrac{45}{6}-\dfrac{3}{6}=\dfrac{8}{6}\)
<=> 9x + 45 - 3 = 8
<=> 9x = 8 + 3 - 45
<=> 9x = -34
<=> x = \(\dfrac{-34}{9}\)
c)\(\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{4}\right)....\left(1+\dfrac{1}{2020}\right)\left(1+\dfrac{1}{2021}\right)\)
\(=\left(\dfrac{1.2}{1.2}+\dfrac{1}{2}\right)\left(\dfrac{1.3}{1.3}+\dfrac{1}{3}\right)...\left(\dfrac{1.2021}{1.2021}+\dfrac{1}{2021}\right)\)
\(=\dfrac{3}{1.2}\cdot\dfrac{4}{1.3}\cdot\cdot\cdot\cdot\dfrac{2022}{1.2021}\)
\(=\dfrac{3.4.5...2022}{\left(1.1.1....1\right)\left(2.3.4...2021\right)}\)
\(=\)\(\dfrac{3.4.5...2022}{2.3.4...2021}\)
\(=\dfrac{2022}{2}=1011\)
\(d\))\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{199}\right)\left(1-\dfrac{1}{200}\right)\)
\(=\left(\dfrac{2}{1.2}-\dfrac{1}{1.2}\right)\left(\dfrac{3}{1.3}-\dfrac{1}{1.3}\right)....\left(\dfrac{200}{1.200}-\dfrac{1}{1.200}\right)\)
\(=\dfrac{1.2.3....199}{\left(1.1.1....1\right).\left(2.3.4....200\right)}\)
\(=\dfrac{1.2.3...199}{2.3.4...200}\)
Nếu mik làm sai mong bạn thông cảm
ý d đáp án là\(\dfrac{1}{200}\) mình quên ghi