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Gọi số thứ nhất phải tìm là X, vậy số thứ 2 sẽ là 27-X.
Do UCLN(X,(27-7))=3 và BCNN(X,(27-X))=60. Do đó ta có X(27-X)=3.60=180. Hay 27X-X^2=180.
X^2-27X+180=0  <=> X^2-15X-12X+180=0 <=> X(X-15)-12(X-15)=0 <=> (X-15)(X-12)=0 Vậy hai số phải tìm là X=12 và X=15 là hai số 12, 15.

4: Chiều rộng của sân là 42:4=10,5(m)

=>Chiều dài của sân cũng là 10,5(m)

Diện tích cái sân là:

\(10,5\times10,5=110,25\left(m^2\right)\)

3:

a: \(137dam^2=1ha37dam^2\)

b: 6km93m=6,093km

c: 2,03 tấn=2 tấn 30kg

d: \(958dm^2=9m^258dm^2\)

A

Bài 22:

1: \(\sqrt{3-\sqrt5}=\frac{\sqrt{6-2\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}}{\sqrt2}=\frac{\sqrt5-1}{\sqrt2}=\frac{\sqrt{10}-\sqrt2}{2}\)

2: \(\sqrt{7+3\sqrt5}\)

\(=\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(3+\sqrt5\right)^2}}{\sqrt2}=\frac{3+\sqrt5}{\sqrt2}=\frac{3\sqrt2+\sqrt{10}}{2}\)

3: \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{17}+1-\sqrt{17}+1\right)-2=\frac{2}{\sqrt2}-2=\sqrt2-2\)

Bài 26:

1: \(\left|3-2x\right|=2\sqrt5\)

=>\(\left[\begin{array}{l}2x-3=2\sqrt5\\ 2x-3=-2\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+2\sqrt5\\ 2x=3-2\sqrt5\end{array}\right.\Rightarrow x=\frac{3\pm2\sqrt5}{2}\)

2: \(\sqrt{x^2}=12\)

=>|x|=12

=>x=12 hoặc x=-12

3: \(\sqrt{x^2-2x+1}=7\)

=>\(\sqrt{\left(x-1\right)^2}=7\)

=>|x-1|=7

=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)

Bài 3: Số $0$

Bài 4: Số $103$

Bài 5: Số $106$

Bài 6: Dư $60$

Bài 7: Số $149$
Bài 8: Số $58$

Bài 9: Số $598$

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

Me: What are you doing on Sunday afternoon?

Friend: I’m not doing anything. / Nothing much.

Me: Do you want to see a movie with me?

Friend: Sure. What movie is it?

Me: It's "The Journey to Mars".

Friend: What time does it start?

Me: It starts at 4 p.m.

Friend: Great! See you then.

Me: What are you doing on Saturday morning?

Friend: I’m free.

Me: Would you like to go to the movies with me?

Friend: Sounds good. What movie do you want to watch?

Me: How about "The Forest Book"?

Friend: What time is it on?

Me: It's at 10 a.m.

Friend: OK, let's meet at the cinema at 9:45.

21 bought (for)

22 spent

23 listening

24 to go

25 reading

26 cleaned

27 went

28 is playing

29 lived

30 received

31 talk

32 met

33 to go

34 stay

35 collecting

36 playing

37 will see

38 play

39 to go

40 to eat