I

1. A ronot is a speacial kind of machine.

2. No, it isn't

3. Worker Robots can help make cars, and explore volcanoes, ...

4. No, it wasn't

II

1. No, there aren't. There are 2 kinds of sports.

2. Team sports require 2 separate teams.

3. Because they want to get the best score.

4. Yes, they are.

31.has not had

32.is

33.sang

34.loves

35.swim

is

sang

loves

swim

Gọi D là giao điểm BM và CN.

Trên cạnh BC lấy điểm E sao cho \(BE=BN\)

Khi đó \(CE=BC-BE=BN+CM-BE=CM\)

Xét hai tam giác BDE và BDN có: 

\(\left\{{}\begin{matrix}BE=BN\\\widehat{DBE}=\widehat{DBN}\left(\text{BM là phân giác}\right)\\BD\text{ chung}\end{matrix}\right.\)  \(\Rightarrow\Delta BDE=\Delta BDN\left(c.g.c\right)\)

\(\Rightarrow\widehat{BDE}=\widehat{BDN}\)

Hoàn toàn tương tự, ta cũng có \(\Delta CDE=\Delta CDM\left(c.g.c\right)\Rightarrow\widehat{CDE}=\widehat{CDM}\)

Mà \(\widehat{BDN}=\widehat{CDM}\) (đối đỉnh) \(\Rightarrow\widehat{BDN}=\widehat{BDE}=\widehat{CDM}=\widehat{CDE}\)

Mà \(\widehat{BDE}+\widehat{CDE}+\widehat{CDM}=180^0\)

\(\Rightarrow3\widehat{BDE}=180^0\Rightarrow\widehat{BDE}=60^0\)

\(\Rightarrow\widehat{CDE}=60^0\)

\(\Rightarrow\widehat{BDC}=\widehat{BDE}+\widehat{CDE}=120^0\)

Theo tính chất tổng 3 góc tổng tam giác:

\(\widehat{BDC}+\widehat{DBC}+\widehat{DCB}=180^0\)

\(\Rightarrow120^0+\dfrac{1}{2}\widehat{B}+\dfrac{1}{2}\widehat{C}=180^0\)

\(\Rightarrow\widehat{B}+\widehat{C}=120^0\)

Do tổng 3 góc trong tam giác ABC bằng 180 độ

\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

\(\Rightarrow\widehat{A}+120^0=180^0\)

\(\Rightarrow A=60^0\)

Bài 5:

a)

\(x^4+x^3+x+1\\ =x^3\left(x+1\right)+\left(x+1\right)\\ =\left(x^3+1\right)\left(x+1\right)\)

b)

\(x^4-x^3-x+1\\ =x^3\left(x-1\right)-\left(x-1\right)\\ =\left(x^3-1\right)\left(x-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\)

c)

\(3x^2-12y^2\\ =\left(\sqrt{3}x\right)^2-\left(\sqrt{12}y\right)^2\\ =\left(\sqrt{3}x-\sqrt{12}y\right)\left(\sqrt{3}x+\sqrt{12}y\right)\\ =\sqrt{3}\left(x-\sqrt{4}y\right).\sqrt{3}\left(x+\sqrt{4}y\right)\\ =3\left(x-\sqrt{4}y\right)\left(x+\sqrt{4}y\right)\)

7:

a: (2x-1)^2-25=0

=>(2x-1)^2=25

=>2x-1=-5 hoặc 2x-1=5

=>2x=6 hoặc 2x=-4

=>x=-2 hoặc x=3

b: 8x^3-50x=0

=>4x^3-25x=0

=>x(4x^2-25)=0

=>x(2x-5)(2x+5)=0

=>x=0 hoặc 2x-5=0 hoặc 2x+5=0

=>x=0;x=5/2;x=-5/2

c: 3x(x-1)+(x-1)=0

=>(x-1)(3x+1)=0

=>x=1 hoặc x=-1/3

d: =>2(x+3)-x(x+3)=0

=>(x+3)(2-x)=0

=>x=-3 hoặc x=2

e: Thiếu vế phải rồi bạn

f: x^3+27+(x+3)(x-9)=0

=>(x+3)(x^2-3x+9)+(x+3)(x-9)=0

=>(x+3)(x^2-3x+9+x-9)=0

=>(x+3)(x^2-2x)=0

=>x(x-2)(x+3)=0

=>\(x\in\left\{0;2;-3\right\}\)

Bài 6: 

\(B=\left(x-1\right)x^2-4x\left(x-1\right)+4\left(x-1\right)\\ =\left(x-1\right)\left(x^2-4x+4\right)\\ =\left(x-1\right)\left(x-2\right)^2\)

Thay x = 3 vào biểu thức B, ta có:

\(B=\left(3-1\right)\left(3-2\right)^2=2\cdot1^2=2\)

Lúc đầu, người bán hàng có: \(50\div\dfrac{2}{5}+1=126\) (trái)

Đ/s:..

Đợi tí, em nhầm :v

Lúc đầu, người bán hàng có: \(\left(50+1\right)\div\left(1-\dfrac{2}{5}\right)=85\)

21+3x−2=48

=>3�−2=48−21=>3x−2=48−21

=>3�−2=27=33=>3x−2=27=33

$=>x-2=3$

$=>x=5$

\(21+3^{x-2}=48\\ \Rightarrow3^{x-2}=27\\ \Rightarrow3^{x-2}=3^3\\ \Rightarrow x-2=3\\ \Rightarrow x=5.\)

thank you các bẹn

101.(-162)+38(-101)+101

= 101.(-162)-38.101+101=101.(-162-38+1)=101.(-199)=-20099