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![](https://rs.olm.vn/images/avt/0.png?1311)
\(2.\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right)\)
= \(\left(6+2\right)\left(3^2+1\right)\left(3^4+1\right)\)
= \(\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\)
= \(\left(3^4-1\right)\left(3^4+1\right)\)
= \(3^8-1\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x + 3)2 - 2(x + 3)(x - 2) + (x - 2)2
= (x + 3 - x + 2)2 = 52 = 25
b) (2x + 5)2 + 2(2x + 5)(3x - 1) + (3x - 1)2
= (2x + 5 + 3x - 1)2 = (5x + 4)2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+1\right)^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)
\(=8x^3+12x^2+6x+1\)
b) \(\left(x-3\right)^3\)
\(=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
Bài 2:
a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)
b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)
c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cái trên của bạn có sai không vậy ?? hình như chỗ -2x phải là -12x
8-12x+6x-x3 =(2-x)3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
c: \(\left(-5x-y\right)^3=-125x^3-75x^2y-15xy^2-y^3\)
h: \(\left(3y-2x^2\right)^3=27y^3-54y^2x^2+36yx^4-8x^6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(\dfrac{3}{4}-\dfrac{1}{2}x\right)\left(\dfrac{3}{4}+\dfrac{1}{2}x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left(x-3+x+2\right)^2\)
\(=\left(2x-1\right)^2\)
Hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b^2\).
b) \(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)
\(=11^2=121\)
Hằng đẳng thức: \(\left(a-b\right)^2=a^2-2ab+b^2\).
a.\(\left(x-3\right)^2+2\left(x-3\right)\left(x+2\right)+\left(x+2\right)^2\)
\(=\left[\left(x-3\right)+\left(x+2\right)\right]^2\)
\(=\left(x-3+x+2\right)^2\)
\(=\left(2x-1\right)^2\)
b.\(\left(x+5\right)^2-\left(2x+10\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left(x+5\right)^2-2\left(x+5\right)\left(x-6\right)+\left(x-6\right)^2\)
\(=\left[\left(x+5\right)-\left(x-6\right)\right]^2\)
\(=\left(x+5-x+6\right)^2\)
\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(=\left(6+2\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\)
\(=3^8-1\)
cảm ơn bn