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Theo đầu bài ta có:
\(2x\left(x-\frac{1}{7}\right)=0\)
\(\Rightarrow\hept{\begin{cases}2x=0\\x-\frac{1}{7}\end{cases}=0}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{1}{7}\end{cases}}\)

\(2x.\left(x-\frac{1}{7}\right)=0\Rightarrow2x=0\)hoặc \(x-\frac{1}{7}=0\)

\(\Rightarrow x=0\)hoặc \(x=\frac{1}{7}\)

\(x^4-3x^3-6x^2+3x+1\)

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

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

đặt \(a=x^2-1\) khi đó biểu thức trở thành

\(a^2-3ax-4x^2\)

\(=a^2+ax-4ax-4x^2\)

\(=\left(a+x\right)\left(a-4x\right)\)

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

a) xét tg ABC và tg DEC có

BC = CE (gt)

AC = CD (gt)

Góc BCA = góc ECD (đối đỉnh)

=> tg ABC = tg DEC (c-g-c)

b) vì tg ABC = tg DEC (cmt)

=> góc BAC = góc CDE ( góc tương ứng)

Mà  góc BAC = 90° (gt)

=> góc CDE = 90°

c) xét tg ACE và tg DCB có:

Góc BCD = góc ACE (đối đỉnh)

 AC = CD ( gt)

CE = CB (gt)

=> tg ACE = tg BDC (c-g-c)

=> góc CBD = góc CEA ( góc tương ứng)

Mà hai góc này ở vị trí soi le trong

=> AE // BD 

Vì tg ACE = tg BDC (cmt)

=> AE = BD ( cạnh tương ứng)

Bài 2 

a, bạn tự vẽ 

b, Hoành độ giao điểm tm pt 

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

\(\Delta'=1-3.2=-5< 0\)

Vậy pt vô nghiệm hay (d) ko cắt (P)

2 she had got the price of the contest, she would be happy now

3 I had gone to the concert last night ,I would be tired

4 John had finished all his homework, he could see a movie now

5 we had listened to my mother, we wouldn't be in trouble right now

Ex5

1 you arrive at the offcie earlier than I do, please turn on the air-conditioner

2 Were it not snowy, the children would go to school

3 he not died so young, he would be a famous musician

4 Were I you, I would take care of my car

5 not very bad-tempered, his wife would give him soon are marrage

6 Had she not got married at such an early age, he would be at university