Độ dãn của lò xo:

\(F_{đh}=k\cdot\Delta l\Rightarrow\Delta l=\dfrac{F_{đh}}{k}=\dfrac{2}{100}=0,02\)m=2cm

B ... Thì lo xo dài 35cm tính f2

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

M cảm ơn b nhé

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

\(=\left(\dfrac{1}{49}-\dfrac{1}{9}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49}\right)\cdot...\cdot\left(\dfrac{1}{49}-\dfrac{1}{49^2}\right)=0\)

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) 

 \(\\ \)

B ơi, chỉ giúp mik bài 3 đc k ạ! Mik cảm ơn b

6: Qua C, kẻ tia CM nằm giữa hai tia CA và CD sao cho CM//DE//AB

CM//DE

=>\(\hat{MCD}=\hat{CDE}\) (hai góc so le trong)

=>\(\hat{MCD}=60^0\)

Ta có: tia CM nằm giữa hai tia CA và CD

=>\(\hat{ACM}+\hat{DCM}=\hat{ACD}\)

=>\(\hat{ACM}=110^0-60^0=50^0\)

Ta có: CM//AB

=>\(\hat{BAC}=\hat{ACM}\) (hai góc so le trong)

=>\(\hat{BAC}=50^0\)

BÀi 5:

Qua B, kẻ tia BM nằm giữa hai tia BA và BC sao cho BM//Ax

BM//Ax

=>\(\hat{xAB}+\hat{ABM}=180^0\) (hai góc trong cùng phía)

=>\(\hat{ABM}=180^0-120^0=60^0\)

Ta có: tia BM nằm giữa hai tia BA và BC

=>\(\hat{ABM}+\hat{CBM}=\hat{ABC}\)

=>\(\hat{CBM}=140^0-60^0=80^0\)

Ta có: \(\hat{CBM}+\hat{BCy}=80^0+100^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên BM//Cy

mà BM//Ax

nên Ax//Cy