Bài 1:

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

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

\(c,3a^2-6ab+3b^2-12c^2=3\left(a-b\right)^2-12c^2=3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)

\(d,x^2-25+y^2+2xy=\left(x-y\right)^2-25=\left(x-y-5\right)\left(x-y+5\right)\)

Bài 1:

\(e,a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)

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

\(g,x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)

\(h,x^2\left(x-1\right)+16\left(1-x\right)=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

1) \(=\left(9x^2-25y^2\right)-\left(6x-10y\right)=\left(3x-5y\right)\left(3x+5y\right)-2\left(3x-5y\right)=\left(3x-5y\right)\left(3x+5y-2\right)\)

2) \(=9x^2y^2-\left(x^2-2xy+y^2\right)=9x^2y^2-\left(x-y\right)^2=\left(3xy-x+y\right)\left(3xy+x-y\right)\)

1) \(\left(x+\dfrac{1}{3}\right)^3=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3\)

\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)

2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)

\(=8x^3+12x^2y^2+6xy^4+y^6\)

3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)

\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)

4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36x^2y^2-8y^3\)

5) \(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\left(\dfrac{2}{3}x^2\right)^3-3.\left(\dfrac{2}{3}x^2\right)^2.\dfrac{1}{2}y+3.\dfrac{2}{3}x^2.\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\)

\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)

6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

\(=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)

7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

8) \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

\(=x+1\)

9) \(\left(x-3\right)\left(x^2+3x+9\right)\)

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

\(=x^3-3^3\)

\(=x^3-27\)

10) \(\left(x-2\right)\left(x^2+2x+4\right)\)

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

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

\(=x^3-8\)

11) \(\left(x+4\right)\left(x^2-4x+16\right)\)

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

\(=x^3+4^3\)

\(=x^3+64\)

12) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)

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

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

\(=x^3-27y^3\)

13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)

\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)

\(=x^6-\dfrac{1}{27}\)

14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

1) \(\dfrac{IM}{IA}=\dfrac{DM}{AB}=\dfrac{CM}{AB}=\dfrac{KM}{KB}\)

Từ đó suy ra được IK // AB // CD.

2) \(\dfrac{IE}{AB}=\dfrac{DI}{DB}=\dfrac{KM}{KB}=\dfrac{IK}{AB}\) --> \(IE=IK\) nên I là trung điểm EK.

\(\dfrac{IK}{AB}=\dfrac{MK}{MB}=\dfrac{CF}{CB}=\dfrac{KF}{AB}\) --> \(IK=KF\) nên K là trung điểm IF.

`# \text {Ryo}`

`2,`

`a)`

`x^3 -9x^2 + 27x - 27`

`= (x)^3 - 3*x^2 * 3 + 3*x*3^2 - (3)^3`

`= (x - 3)^3`

`b)`

`- (x^3)/8 + 3/4x^2 - 3/2x + 1`

`= - ( (x^3)/8 - 3/4x^2 + 3/2x - 1)`

`= - [ (x/2)^3 - 3*(x/2)^2 * 1 + 3*x/2*1^2 - 1^3]`

`= - (x/2 - 1)^3`

`c)`

Phiền bạn ghi lại đề giúp mình với ạ! Số mũ của biến 3 số sau mình kh đọc được.

`3,`

`a)`

`A = x^3 - 6x^2 + 12x - 8`

`= (x)^3 - 3*x^2*2 + 3*x*2^2 - (2)^3`

`= (x - 2)^3`

`b)`

`B = 1 - (3x)/2 + (3x^2)/4 - (x^3)/8` phải k c? (Mình thấy biến phần cuối hơi mờ).

`= 1^3 - 3*1^2*x/2 + 3* 1 * (x/2)^2 - (x/2)^3`

`= (1 - x/2)^3`

__

Cả bài 2 và 3, bạn sử dụng CT:

`A^3 - 3A^2B + 3AB^2 - B^3 = (A - B)^3`

2, c là x⁶ - \(\dfrac{3}{2}x^4y+\dfrac{3}{4}x^2y^2-\dfrac{1}{8}y^3\)
còn 3 b là đúm rùi ạ

Xét tứ giác AMDN có 

AM//DN

AN//DM

Do đó: AMDN là hình bình hành

Suy ra: Hai đường chéo AD và MN cắt nhau tại trung điểm của mỗi đường

mà I là trung điểm của AD

nên I là trung điểm của MN

a: Ta có: \(\left(3x-1\right)^2-\left(3x+4\right)\left(3x-4\right)=32\)

\(\Leftrightarrow9x^2-6x+1-9x^2+16=32\)

\(\Leftrightarrow-6x=15\)

hay \(x=-\dfrac{5}{2}\)

b: Ta có: \(\left(4x+3\right)^2-\left(4x-1\right)\left(4x+1\right)=-14\)

\(\Leftrightarrow16x^2+24x+9-16x^2+1=-14\)

\(\Leftrightarrow24x=-24\)

hay x=-1

Vì \(BD\text{//}AD;CD\text{//}AB\) nên ABDC là hình bình hành

Mà \(\widehat{BAC}=90^0\) nên ABDC là hình chữ nhật