\(M=\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+...+\frac{3}{2021.2022}\)

    \(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2021.2022}\right)\)

    \(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2021}-\frac{1}{2022}\right)\)

    \(=3\left(1-\frac{1}{2022}\right)\)

    \(=3.\frac{2021}{2022}\)

    \(=\frac{6063}{2022}\)

Câu 5: Đặt \(\frac{x}{z}=\frac{z}{y}=k\)

=>z=yk; x=zk=yk*k=yk^2

\(\frac{x^2+z^2}{y^2+z^2}=\frac{\left(yk^2\right)^2+\left(yk\right)^2}{y^2+\left(yk\right)^2}=\frac{y^2k^4+y^2k^2}{y^2+y^2k^2}=\frac{y^2k^2\left(k^2+1\right)}{y^2\left(1+k^2\right)}=k^2\)

\(\frac{x}{y}=\frac{yk^2}{y}=k^2\)

Do đó: \(\frac{x^2+z^2}{y^2+z^2}=\frac{x}{y}\)

Câu 4:

a: Ta có: \(\hat{x^{\prime}Mz}=\hat{MNy^{\prime}}\left(=120^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên xx'//yy'

b: Qua B, kẻ tia BE nằm giữa hai tia BA và BC sao cho BE//AM//CN

BE//AM

=>\(\hat{ABE}=\hat{xAB}\) (hai góc so le trong)

=>\(\hat{ABE}=35^0\)

BE//CN

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

=>\(\hat{EBC}=180^0-125^0=55^0\)

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

=>\(\hat{ABC}=\hat{ABE}+\hat{CBE}=55^0+35^0=90^0\)

=>BA⊥BC

Gọi vận tốc dự kiến là x(km/h)

(Điều kiện: x>15)

Vận tóc trên 1/4 quãng đầu tiên là x-15(km/h)

Vận tốc trên 3/4 quãng đường còn lại là x+10(km/h)

1/4 quãng đường=80*1/4=20km

3/4 quãng đường=80*3/4=60(km)

Thời gian ô tô dự định sẽ đi hết quãng đường là:

\(\frac{80}{x}\) (giờ)

Thời gian ô tô đi hết 20km đầu tiên là \(\frac{20}{x-15}\) (giờ)

Thời gian ô tô đi hết 60km còn lại là \(\frac{60}{x+10}\) (giờ)

Ô tô đến B đúng dự kiến nên ta có:

\(\frac{80}{x}=\frac{20}{x-15}+\frac{60}{x+10}\)

=>\(\frac{4}{x}=\frac{1}{x-15}+\frac{3}{x+10}\)

=>\(\frac{4}{x}=\frac{x+10+3x-45}{\left(x-15\right)\left(x+10\right)}\)

=>\(\frac{4}{x}=\frac{4x-35}{\left(x-15\right)\left(x+10\right)}\)

=>x(4x-35)=4(x-15)(x+10)

=>\(4x^2-35x=4\left(x^2+10x-15x-150\right)=4\left(x^2-5x-150\right)\)

=>\(4x^2-35x=4x^2-20x-600\)

=>-35x=-20x-600

=>-15x=-600

=>x=40(nhận)

Thời gian ô tô dự định sẽ đi hết quãng đường là:

80/40=2(giờ)

\(\overline{ab}+\overline{ba}=11\left(a+b\right)\)

11 là số nguyên tố để 11(a+b) là số chính phương thì a+b=11

\(\Rightarrow\overline{ab}=\left\{29;38;47;56;65;74;83;92\right\}\)

1. John was sad yesterday.

2. The birds were in the cage.

3. The children were in the class.

4. My cat was on the chair.

5. I was at the theater yesterday.

6. It was foggy this morning.

7. They were at home last Sunday.

8. I was very happy yesterday.

9. There were a lot of balloons.

10. You were at the park last night.

các bài khác đâu ạ gt đâu r bạn ạ

1 his illness, he cannot come

2 her busyness, she couldn't help us

3 his illness, he tries to go to school on time

4 the bad weather, we tried to finish the work on the road

5 the bad weather, we got to the station late

6 the old house, she liked it

7 not wearing any shoes, Carol ran outside to see what was happening

8 being afraid of flying, Fiona had to get on the plane

\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)

\(x^2-y^2-6x+9=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x+y-3\right)\left(x-y-3\right)\)

Cảm ơn bạn nha :>>