Tìm nghiệm của đa thức -10x3 - 1x -3 = 0
Giải hộ mik vs ạ mik đang cần gấp,cảm ơn nhiều ạ!
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1 are
2 am
3 is
4 are
5 are
6 are
7 is
8 is
9 is
10 are
IV
1 is writing
2 are losing
3 is having
4 is staying
5 am not lying
6 is always using
7 are having
8 Are you playing
9 are not touching
10 Is - listening
11 Is- winning
12 am not staying
13 is not working
14 is not reading
15 isn't raining
16 am not listening
17 Are they making
18 Are you doing
19 Is - sitting
20 is - doing
21 are-putting
22 are-wearing
23 is-studying
2, am
3, is
4,are
5,are
6,are
7,is
8,is
9,is
10,are
IV
1,2,7 OK
3,is having
4,has stayed
5,am not lying
6,always uses
8,Are-playing
9,not to touch
10,Is-listening
11,Are-winning
12,am not staying
13,isn't working
14,isn't reading
15,isn't raining
16,am not listening
17,Are-making
18,Are-doing
19,Is-sitting
20,is-doing
21,do-putting
22,do-wear
23,is-studying
Kẻ Bz//Ax
Ta có: Ax//Bz
\(\Rightarrow\widehat{BAx}=\widehat{ABz}=30^0\)(so le trong)
\(\Rightarrow\widehat{zBC}=\widehat{ABC}-\widehat{BAx}=90^0-30^0=60^0\)
Ta có: \(\widehat{zBC}+\widehat{BCy}=60^0+120^0=180^0\)
Mà 2 góc này là 2 góc trong cùng phía
=> Bz//Cy
Mà Bz//Ax
=> Ax//Cy
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x=t\)
\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)
\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)
Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
vâng ạ
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