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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
