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a) ta có : \(-x^3+3x^2-3x+1=1-3.1^2.x+3.1.x^2-x^3=\left(1-x\right)^3\)
b) ta có : \(64-48x+12x^2-x^3=4^3-3.4^2x+3.4.x^2-x^3=\left(4-x\right)^3\)
Ta có: a/ -x3+3x2-3x+1 = -(x3-3x2+3x-1)
= -(x-1)3
b/ 64-48x+12x2-x3 = 43-3.42.x+3.a.x2-x3
= (4-x)3
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)
\(=\left(x+4\right)^3\)
b) Ta có: \(x^3-12x^2+48x-64\)
\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)
\(=\left(x-4\right)^3\)
c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=\left(2x+y\right)^3\)
d)Sửa đề: \(x^3-3x^2+3x-1\)
Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
e) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mk ko hỉu cái đề của bn: Dạng 4,5: Lập phương của 1 tổng và lập phương của một hiệu ♥
Có phải bằng Dạng 4,5: Lập phương của 1 tổng và lập phương của một hiệu là yo
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a) \(8-12x+6x^2-x^3\)
\(=-x^3+8+6x^2-12x\)
\(=-\left(x^3-2^3\right)+6x\left(x-2\right)\)
\(=-\left(x-2\right)\left(x^2+2x+4\right)+6x\left(x-2\right)\)
\(=\left(x-2\right)\left(-x^2-2x-4+6x\right)\)
\(=\left(x-2\right)\left(-x^2+4x-4\right)\)
\(=-\left(x-2\right)\left(x-2\right)^2\)
\(=-\left(x-2\right)^3\)
b) \(48x+64+x^3+12x^2\)
\(=x^3+3.4.x^2+3.x.4^2+4^3\)
\(=\left(x+4\right)^3\)
c) \(-9y^2+y-\dfrac{1}{27}+27y^3\)
\(=27y^3-9y^2+y-\dfrac{1}{27}\)
\(=\left(3y\right)^3-3.\left(3y\right)^2.\dfrac{1}{3}+3.3y.\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3\)
\(=\left(3y-\dfrac{1}{3}\right)^3\)
d) \(8x^3+150x-125-60x^2\)
\(=8x^3-60x^2+150x-125\)
\(=\left(2x\right)^3-3.\left(2x\right)^2.5+3.2x.5^2-5^3\)
\(=\left(2x-5\right)^3\)
a, \(8-12x+6x^2-x^3=-\left(x^3-6x^2+12x-8\right)\)
\(=-\left(x^3-2x^2-4x^2+8x+4x-8\right)\)
\(=-\left(x-2\right)^3\)
b, \(48x+64+x^3+12x^2=x^3+4x^2+8x^2+32x+16x+24\)
\(=\left(x+4\right)^3\)
c, \(-9y^2+y-\dfrac{1}{7}+27y^3\)
(sai đề)
d, \(8x^3+150x-125-60x^2=8x^3-20x^2-40x^2+100x+50x-125\)
\(=4x^2\left(2x-5\right)-20x\left(2x-5\right)+25\left(2x-5\right)\)
\(=\left(2x-5\right)\left(4x^2-20x+25\right)=\left(2x-5\right)\left(2x-5\right)^2\)
\(=\left(2x-5\right)^3\)
Chúc bạn học tốt!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3\)
\(x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3.
a, (2y- 1)3= (2y)3-3.(2y)2.1+3.2y.12-13
= 8y3-12y2+6y-1
b, (3x2+2y)3=(3x2)3+3.(3x2)2.2y+3.3x2.(2y)2+13
=27x6+54x4y+36x2y2+1
c, ( 1/3x-2)3=(1/3x)3-3.(1/3x)2.2+3.1/3x.22-23
=1/27x3-2/3x2+4x-8
4.
a, -x3+3x3-3x+1=1-3x+3x3-x3
=1-3.12.x+3.1.x3-x3
=(1-x)3
b,64-48x+12x2-x3=43-3.42.x+3.4.x2-x3
=(4-x)3
Bài 3 Tính:
\(a\)) \(\left(2y-1\right)^3=2y^3-3.\left(2y\right)^2.1+3.2y.1^2-1^3\)
\(=2y^3-12y^2+6y-1\)
b)\(\left(3x^2+2y\right)^3\)
\(=\left(3x^2\right)^3=3.\left(3x^2\right)^2.2y+3.\left(3x^2\right).\left(2y\right)^2+\left(2y\right)^3\)
\(=27x^8+3.9x^4.2+9x^2.4y+8y^3\)
\(=27x^8+54x^4+36x^2y+8y^3\)
c)\(\left(\dfrac{1}{3}x-2\right)^3\)
\(=\left(\dfrac{1}{3}x\right)^3-3.\left(\dfrac{1}{3}x\right)^2.2+3.\dfrac{1}{3}x.2^2-2^3\)
\(=\dfrac{1}{27}x^3-3.\dfrac{1}{9}x^2.2+x.2^2-8\)
\(=\dfrac{1}{27}x^3-\dfrac{2}{3}x^2+4x-8\)