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.

a) x-15+124=200

x-15=200-124

x-15=76

x=76+15

x=91

Vậy x=91.

b) 3x-21=3⁶:3⁴

3x-21=3²

3x-21=9

3x=9+21

3x=30

x=30:3

x=10

Vậy x=10.

c) x-48:16=21

x-3=21

x=21+3

x=24

Vậy x=24.

a,(x-15)+124=200

   x - 15         = 200 - 124

  x - 15          = 76

  x                  = 76 + 15

  x                  = 91.

Vậy x = 91.

b,3x-21=3⁶:3⁴

   3x - 21 = 3²

    3x - 21 = 9

     3x       = 9 + 21

     3x       = 30

      x         = 30 : 3

      x         = 10.

Vậy x = 10.

c,x-48:16=21

    x - 3    = 21

     x         = 21 + 3

     x         = 24.

Vậy x = 24

# HOK TỐT #

31.has not had

32.is

33.sang

34.loves

35.swim

is

sang

loves

swim

\(2x+12-3x=-21\)

\(x.\left(2+12-3\right)=-21\)

\(x.11=-21\)

\(x=-\frac{21}{11}\)

\(235-288:\left[4\times\left(48-72\right)\right]\)

\(=235-288:4\times\left(48-72\right)\)

\(=235-72\times\left(-24\right)\)

\(=235-\left(-1728\right)\)

\(=235+1728\)

\(=1963\)

1/ kq: 238.

2/ kq: 33.

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: \(\left(50+1\right)\div\left(1-\dfrac{2}{5}\right)=85\)

Giải:

a) \(\left|48-3x\right|=0\)

\(\Leftrightarrow48-3x=0\)

\(\Leftrightarrow3x=48\)

\(\Leftrightarrow x=\dfrac{48}{3}=16\)

Vậy x = 16.

b) \(\left|-x-7\right|=24\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24\\-x-7=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=31\\-x=-17\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-31\\x=17\end{matrix}\right.\)

Vậy \(x=-31\) hoặc \(x=17\).

c) \(\left|4-x\right|=21\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=21\\4-x=-21\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\\x=25\end{matrix}\right.\)

Vậy \(x=-17\) hoặc \(x=25\).

d) \(\left|x+8\right|+12=0\)

\(\Leftrightarrow\left|x+8\right|=-12\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12\\x+8=-\left(-12\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-20\\x=4\end{matrix}\right.\)

Vậy \(x=-20\) hoặc \(x=4\).

e) \(6-\left|x\right|=2\)

\(\Leftrightarrow\left|x\right|=6-2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

Vậy \(x=4\) hoặc \(x=-4\).

Chúc bạn học tốt!

1. |48 - 3x| = 0.

\(\Leftrightarrow\) 48 - 3x = 0.

\(\Leftrightarrow\) 3x = 48.

\(\Leftrightarrow\) x = \(\dfrac{48}{3}=16.\)

Vậy x = 16.

2. |-x - 7| = 24.

\(\Leftrightarrow\left[{}\begin{matrix}-x-7=24.\\-x-7=-24.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=31.\\-x=-17.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-31.\\x=17.\end{matrix}\right.\)

3. |4 - x| = 21.

\(\Leftrightarrow\left[{}\begin{matrix}4-x=21.\\4-x=-21.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-17.\\x=25.\end{matrix}\right.\)

4. |x + 8| + 12 = 0.

|x + 8| = 0 - 12.

|x + 8| = -12.

\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=-\left(-12\right).\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=-12.\\x+8=12.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-20.\\x=4.\end{matrix}\right.\)

5. 6 - |x| = 2.

|x| = 6 - 2.

|x| = 4.

\(\Leftrightarrow\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=4.\\x=-4.\end{matrix}\right.\)

1. Ta có 2 trường hợp:

TH1: 48-3x=0 <=> 3x=48 <=> x=16

TH2:-48-3x=0