![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)27{{\rm{x}}^3} + 1 = {\left( {3{\rm{x}}} \right)^3} + 1 = \left( {3{\rm{x}} + 1} \right).\left[ {{{\left( {3{\rm{x}}} \right)}^2} - 3{\rm{x}}.1 + {1^2}} \right] = \left( {3{\rm{x}} + 1} \right)\left( {9{{\rm{x}}^2} - 3{\rm{x}} + 1} \right)\)
\(b)64 - 8{y^3} = {4^3} - {\left( {2y} \right)^3} = \left( {4 - 2y} \right)\left[ {{4^2} + 4.2y + {{\left( {2y} \right)}^2}} \right] = \left( {4 - 2y} \right)\left( {16 + 8y + 4{y^2}} \right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(27 + 54x + 36{x^2} + 8{x^3} = {3^3} + {3.3^2}.2x + 3.3.{\left( {2x} \right)^2} + {\left( {2x} \right)^3} = {\left( {3 + 2x} \right)^3}\)
b) \(64{x^3} - 144{x^2}y + 108x{y^2} - 27{y^3} = {\left( {4x} \right)^3} - 3.{\left( {4x} \right)^2}.3y + 3.4x.{\left( {3y} \right)^2} - {\left( {3y} \right)^3} = {\left( {4x - 3y} \right)^3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
c: \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)
\(\dfrac{1}{8}x^3-64=\left(\dfrac{1}{2}x-4\right)\left(\dfrac{1}{4}x^2+2x+16\right)\)
d: \(=\left(2x+5y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x+1\right)^2\\ b,=\left(3x-y\right)^2\\ c,=\left(x-3\right)\left(x+3\right)\\ d,=\left(x+4\right)^3\\ e,=\left(x-2\right)^3\\ f,=\left(x+2\right)\left(x^2-2x+4\right)\\ g,=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
b: \(x^3-\dfrac{1}{8}=\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c: \(8x^3+y^3=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
a) \(\left(x+3\right)\cdot\left(x^2-3x+9\right)\)
b) \(\left(x-\dfrac{1}{2}\right)\cdot\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)\)
c) \(\left(2x+y\right)\cdot\left(4x^2-2xy+y^2\right)\)
![](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
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
=(x-2)3
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
![](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: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)