Ta có \(\frac{x-y}{z}=\frac{3y}{x-z}=\frac{x}{y}=\frac{x-y+3y+x}{z+x-z+y}=\frac{2x+2y}{x+y}=\frac{2\left(x+y\right)}{x+y}=2\)(dãy tỉ số bằng nhau)

=> x = 2y (đpcm)

Khi đó \(\frac{x-y}{z}=2\Leftrightarrow x-y=2z\Rightarrow2y-y=2z\Rightarrow y=2z\)(đpcm)

dpcm là j vậy

1. là,lỏng,luồn

2. nằm , nước ,nến 

3.

4. dõng dạc ,dễ dãi ,bẽn lẽn

Long lanh, luôn luôn, liên lạc.

Nóng nảy, nóng nực, nắng nôi.

Ngủ ngon.

Lẽo đẽo, lẽn bẽn.

Ta có : \(x^5+x+1=0\)

\(\Leftrightarrow x^5-x^2+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x^3-1\right)+\left(x^2+x+1\right)=0\)

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

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^3-x^2+1\right)=0\)

.....

1) She watched that show three times last week

2) I've ever played guitar since I was a teenager

3) The children have made a mess in the kitchen

4) Have you called your mom yet ? 

5) The girl have played tennis at the clup since 2005

6) She has not appied for the job

7) They have made a plan for a starting business,

8) The poems has written by roman poet

9) She has drunk three cups of coffee today

10) Have you been to this place before ? 

1. She has watched that show three times last week.

2. I've ever played the guitar since I was a teenager.

3. The children have made a mess in the kitchen.

4. Have you called your mom yet?

5. The girl have played tennis at the club since 2005.

6. She has not applied for the job.

7. They have made a plan for starting business.

8. The poet has written roman poems.

9. She has drunk three cups of coffee today.

10. Have you been to this place before?

ko đối với mik còn bạn uống đc bạn cứ uống nhen!

sữa dùng để tắm  người ta gọi là sữa tắm ko gọi là sữa xoa( kể cả khi xoa lên người)

vì người ta sợ tốn xà phòng hoặc người ta thương nó hay sợ tiếc của nên ko dám dùng hay nói cách khác là sợ!

câu hỏi này phù hợp với đố vui đấy. Mình nghĩ bạn ra câu hỏi ở đố vui chứ không phải là Tiếng Anh.

ok

\(M=\frac{17}{5}\cdot\frac{-31}{125}\cdot\frac{1}{2}\cdot\frac{10}{17}\cdot\frac{-1}{2^3}\)

\(M=\frac{17}{5}\cdot\frac{-31}{125}\cdot\frac{1}{2}\cdot\frac{10}{7}\cdot\frac{-1}{8}\)

\(M=\left(\frac{17}{5}\cdot\frac{10}{17}\cdot\frac{1}{2}\right)\cdot\frac{-31}{125}\cdot\frac{-1}{8}\)

\(M=1\cdot\frac{31}{1000}=\frac{31}{1000}\)

\(P=\frac{6}{7}\cdot\frac{8}{13}+\frac{6}{9}\cdot\frac{9}{7}-\frac{3}{13}\cdot\frac{6}{7}=\frac{6}{7}\cdot\frac{8}{13}+\frac{6}{7}\cdot1-\frac{3}{13}\cdot\frac{6}{7}\)

\(=\frac{6}{7}\left(\frac{8}{13}+1-\frac{3}{13}\right)=\frac{6}{7}\left(\frac{8}{3}+\frac{13}{13}-\frac{3}{13}\right)=\frac{6}{7}\cdot\frac{18}{13}=\frac{108}{91}\)

x⁵+x+1

x⁵-x²+x²+x+1

x²(x³-1)+(x²+x+1)

x²(x-1)(x²+x+1)+(x²+x+1)

[x²(x-1)](x²+x+1)

a) ĐKXĐ : \(x\ne0\);\(x\ne2;-2\)

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

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

       =\(\frac{x+2+2x+x-2}{\left(x+2\right)\left(x-2\right)}.\frac{2-x}{x}\)

       =\(\frac{4x}{\left(x+2\right)\left(x-2\right)}.\frac{-\left(x-2\right)}{x}\)

       =  \(\frac{-4}{x+2}\)

b) Ta có : \(2x^2+x=0\)

        \(\Leftrightarrow x\left(2x+1\right)=0\)

        \(\Leftrightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{-1}{2}\end{cases}}\left(tm\right)\)

Để A = -1/2 thì 

\(\Leftrightarrow\frac{-4}{x+2}=\frac{-1}{2}\)

\(\Leftrightarrow-\left(x+2\right)=-8\)

\(\Leftrightarrow x+2=8\)

\(\Leftrightarrow x=6\)

c) Để A =0,5 thì 

\(\frac{-4}{x+2}=0,5\)

\(\Leftrightarrow-8=x+2\)

\(\Leftrightarrow x=-10\)

d) Để A \(\inℤ\)thì

\(-4⋮x+2\)

\(\Leftrightarrow x+2\inƯ\left(-4\right)\)

\(\Leftrightarrow x+2\in\left\{1;2;4;-1;-2;-4\right\}\)

Lập bảng giá trị 

Mà \(x\ne0\)và \(x\ne2;-2\)

\(\Rightarrow x\in\left\{-1;-3;-4;-6\right\}\)

\(=\frac{\sqrt{5}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}-\sqrt{3}\right)}\)

\(=\frac{\sqrt{5}}{2}\)