\(\frac{1}{b}+\frac{1}{c}=\frac{1}{d}\Leftrightarrow\frac{b+c}{bc}=\frac{1}{d}\Leftrightarrow d=\frac{bc}{b+c}\)

Ta có

\(HD\perp AB;AC\perp AB\) => HD//AC \(\Rightarrow\frac{BD}{BC}=\frac{HD}{AC}=\frac{d}{b}\Rightarrow d=\frac{b.BD}{BC}\) (*)

Xét tg ABC có AD là phân giác của \(\widehat{A}\) nên

\(\frac{BD}{AB}=\frac{CD}{AC}\) (Trong tam giác, đường phân giác của một góc chia cạnh đối diện thành hai đoạn tỉ lệ với hai cạnh kề của hai đoạn ấy)

\(\Rightarrow\frac{BD}{c}=\frac{CD}{b}=\frac{BD+CD}{b+c}=\frac{BC}{b+c}\Rightarrow BC=\frac{BD.\left(b+c\right)}{c}\) Thay vào (*)

\(d=\frac{b.BD}{\frac{BD.\left(b+c\right)}{c}}=\frac{b.BD.c}{BD.\left(b+c\right)}=\frac{bc}{b+c}\Leftrightarrow\frac{1}{b}+\frac{1}{c}=\frac{1}{d}\left(dpcm\right)\)

30. What’s your date of birth?

When is your birthday?

31. Whose documents are these?

Who is the owner of these documents?  

32. how much does it cost?

What is the price of it ?

How much is it?

33. 1. It is a terrible dish.

⇒ What a terrible dish !

34. These boys are very naughty.

⇒ What boys naughty.

35. His drawing is very perfect.

⇒ How perfect  his drawing is !

36. He draws a perfect picture!

⇒ What  a perfect picture!

37. He writes very carelessly!

⇒ How carelessly he write is !

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow x\left(2x^2+10x-x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-6x=8\Leftrightarrow x=-\frac{4}{3}\)

\(x\left(2x-1\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^2-x\right)\left(x+5\right)-\left(2x^2+1\right)\left(x+4,5\right)=3,5\)

\(\Leftrightarrow\left(2x^3+9x^2-5x\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\)

\(\Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\)

\(\Leftrightarrow-6x-4,5=3,5\)

\(\Leftrightarrow-6x=3,5+4,5\)

\(\Leftrightarrow-6x=8\)

\(\Leftrightarrow x=-\frac{8}{6}=-\frac{4}{3}\)

TRẢ LỜI:

Thực hiện phép chia:

2x3 – 3x2 + x + a chia hết cho x + 2

⇔ số dư = a – 30 = 0

⇔ a = 30.

Cách 2: Phân tích 2x3 – 3x2 + x + a thành nhân tử có chứa x + 2.

2x3 – 3x2 + x + a

= 2x3 + 4x2 – 7x2 – 14x + 15x + 30 + a – 30

(Tách -3x2 = 4x2 – 7x2; x = -14x + 15x)

= 2x2(x + 2) – 7x(x + 2) + 15(x + 2) + a – 30

= (2x2 – 7x + 15)(x + 2) + a – 30

2x3 – 3x2 + x + a chia hết cho x + 2 ⇔ a – 30 = 0 ⇔ a = 30

x thôi sao 4x

đa thức  2x³ - x² + 4x + a chia hết cho đa thức x + 2 khi  - 5x mũ 14x+a - 14x+28=0

                                                                                                                a  -      28  = 0

                                                                                                                a               =0 +28

                                                                                                                a                =28

 2x³ - x² + 4x + a : x+2 =2x mũ 2 -5x+14  

good luck!

Ta có : 3y² + x² + 2xy + 2x + 6y + 3 = 0

=> (x² + 2xy + y²) + (2x + 2y) + 1 + (2y² + 4y + 2) = 0

=> (x + y)² + 2(x + y) + 1 + 2(y² + 2y + 1) = 0 

=> (x + y + 1)² + 2(y + 1)² = 0

=> \(\hept{\begin{cases}x+y+1=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Vậy x = 0 ; y = -1 là giá trị cần tìm

\(3x^2+x^2+2xy+2x+6y+3=0\)

\(\left(x^2+2xy+y^2\right)+\left(2y^2+4y+2\right)+\left(2y+2x\right)+1=0\)

\(\left(x+y\right)^2+2\left(y^2+2y+1\right)+2\left(x+y\right)+1=0\)

\(\left(x+y\right)^2+2\left(y+1\right)^2+2\left(x+y\right)+1=0\)

\(\left(x+y+1\right)^2+2\left(y+1\right)^2=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+1=0\\y+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\y=-1\end{cases}}}\)

3y^2 + x^2 + 2xy + 2x + 6y + 3 = 0

<=> (x^2 + 2xy + y^2) + 2y^2 + 2x + 6y + 3 = 0`

<=> (x + y)^2 + 2(x + y) + 1 + 2y^2 + 4y + 2 = 0`

`<=> (x + y + 1)^2 + 2(y + 1)^2 = 0`

<=> {x + y + 1 = 0

      {y + 1 = 0

<=> {x = 0

       {y = -1

bài này dễ quá

https://olm.vn/hoi-dap/detail/329012981556.html

\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}=\frac{1+x+1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}=\frac{2\left(1+x^2\right)+2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}=\frac{4\left(1+x^4\right)+4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)

\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}=\frac{8\left(1+x^8\right)+8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)

\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}=\frac{16\left(1+x^{16}\right)+16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}=\frac{32}{1-x^{32}}\)