o: x^4+x^3+x^2-1

=x^3(x+1)+(x-1)(x+1)

=(x+1)(x^3+x-1)

q: \(=\left(x^3-y^3\right)+xy\left(x-y\right)\)

=(x-y)(x^2+xy+y^2)+xy(x-y)

=(x-y)(x^2+2xy+y^2)

=(x-y)(x+y)^2

s: =(2xy)^2-(x^2+y^2-1)^2

=(2xy-x^2-y^2+1)(2xy+x^2+y^2-1)

=[1-(x^2-2xy+y^2]+[(x+y)^2-1]

=(1-x+y)(1+x-y)(x+y-1)(x+y+1)

u: =(x^2-y^2)-4(x+y)

=(x+y)(x-y)-4(x+y)

=(x+y)(x-y-4)

x: =(x^3-y^3)-(3x-3y)

=(x-y)(x^2+xy+y^2)-3(x-y)

=(x-y)(x^2+xy+y^2-3)

z: =3(x-y)+(x^2-2xy+y^2)

=3(x-y)+(x-y)^2

=(x-y)(x-y+3)

o) \(x^4+x^3+x^2-1\)

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

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

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

q) \(x^3+x^2y-xy^2-y^3\)

\(=\left(x^3+x^2y\right)-\left(xy^2+y^3\right)\)

\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-y^2\right)\)

\(=\left(x+y\right)^2\left(x-y\right)\)

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

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

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

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

\(=-\left(x-y-1\right)\left(x-y+1\right)\left(x+y+1\right)\left(x+y-1\right)\)

u) \(x^2-y^2-4x-4y\)

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

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

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

x) \(x^3-y^3-3x+3y\)

\(=\left(x^3-y^3\right)-\left(3x-3y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2-3\right)\)

z) \(3x-3y+x^2-2xy+y^2\)

\(=\left(3x-3y\right)+\left(x^2-2xy+y^2\right)\)

\(=3\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x-y\right)\left(3+x-y\right)\)

She left the note on the door because the house is empty

Since Lan was careless, she lost the game

D

\(lim\dfrac{2\sqrt{7n^2-2n}}{3n+2}=lim\dfrac{2\sqrt{n^2\left(7-\dfrac{2}{n}\right)}}{3n+2}=lim\dfrac{2n\sqrt{7-\dfrac{2}{n}}}{n\left(3+\dfrac{2}{n}\right)}\)

\(=lim\dfrac{2\sqrt{7-\dfrac{2}{n}}}{3+\dfrac{2}{n}}=\dfrac{2\sqrt{7}}{3}\) \(=\dfrac{a\sqrt{7}}{b}\) 

Suy ra : a/b = 2/3 => a - b = -1 

1 phút = 60s

- Chu kì: T = \(\dfrac{t}{n}\)= \(\dfrac{60}{4}\)= 15 (s)

- Tần số: f = \(\dfrac{1}{T}\)=\(\dfrac{1}{15}\)(Hz)

- Tốc độ góc: T = \(\dfrac{2\pi}{\omega}\) \(\Rightarrow\) \(\omega=\dfrac{2\pi}{T}\)\(=\) \(\dfrac{2\pi}{15}=\dfrac{2}{15}\pi\) (rad/s)

- Tốc độ dài: \(\upsilon=r\omega\) ⇔ \(\upsilon=\dfrac{3}{2}.\dfrac{2}{15}\pi\) = \(\dfrac{1}{5}\pi\) (m/s)

- Gia tốc hướng tâm: \(a_{ht}=r.\omega^2\Leftrightarrow a_{ht}=\dfrac{3}{2}.\left(\dfrac{2}{15}\pi\right)^2\)\(\simeq\) 0,263 (m/\(s^2\))

\(72.\)

Xét tam giác AHC vuông tại H ta có:

\(\sin C=\dfrac{AH}{AC}\) (Tỉ số lượng giác).

\(\Rightarrow\sin30^o=\dfrac{AH}{40}.\Rightarrow AH=20.\)

Xét tam giác AHB vuông tại H:

\(AB^2=AH^2+BH^2\left(Pytago\right).\)

\(\Rightarrow29^2=20^2+BH^2.\\ \Leftrightarrow BH^2=29^2-20^2.\\ \Rightarrow BH=21.\)

Ko cần hình đâu ạ

Xét tam giác vuông AHC có:

Góc HCA = 30 độ => Tam giác AHC là tam giác nửa đều

=> AH=\(\dfrac{1}{2}\) AC ( Tính chất tam giác nửa đều )

=> AH = 20 

Xét tam giác AHB có:

( Định lý Py ta go )

=> BH= 21

Vậy BH=21

còn hình bn tự vẽ nhé

1 B
2C
3A
4B
5A
7D
8A
9A
10C