![](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) x⁶ + y⁶ = (x²)³ + (y²)³
= (x² + y²)(x⁴ - x²y² + y⁴)
b) x⁶ - y⁶
= (x³)² - (y³)²
= (x³ - y³)(x³ + y³)
= (x - y)(x² + xy + y²)(x + y)(x² - xy + y²)
![](https://rs.olm.vn/images/avt/0.png?1311)
x 6 - y 6 = x 3 2 - y 3 2 = x 3 + y 3 x 3 - y 3 = x + y x 2 - x y + y x - y x 2 + x y + y 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
x 6 - y 6 = x 3 2 - y 3 2 = x 3 + y 3 x 3 - y 3 = x + y x 2 - x y + y 2 x - y x 2 + x y + y 2
Đáp án cần chọn là : C
![](https://rs.olm.vn/images/avt/0.png?1311)
Tròn đã làm bằng cách:
\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3\)
\(=\left(x^2+y^2\right)\left[\left(x^2\right)^2-x^2\cdot y^2+\left(y^2\right)^2\right]\)
\(=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
\({x^6} + {y^6} = {\left( {{x^2}} \right)^3} + {\left( {{y^2}} \right)^3} = \left( {{x^2} + {y^2}} \right)\left[ {{{\left( {{x^2}} \right)}^2} - {x^2}.{y^2} + {{\left( {{y^2}} \right)}^2}} \right] = \left( {{x^2} + {y^2}} \right)\left( {{x^4} - {x^2}{y^2} + {y^4}} \right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
P = x6 + y6 = (x2 + y2)(x4 - x2 y2 + y4)
= (x2 + y2)2 - 3x2 y2 \(\ge1-3×\frac{\left(x^2+y^2\right)^2}{4}=1-\frac{3}{4}=\frac{1}{4}\)
Đạt được khi x2 = y2 = \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(8^3yz+12^2yz+6xyz+yz\)
\(=512yz+144yz+6xyz+yz\)
\(=yz\left(512+14+6x+1\right)\)
\(=yz\left(527+6x\right)\)
$---$
b, \(81x^4\left(z^2-y^2\right)-z^2+y^2\)
\(=81x^4\left(z^2-y^2\right)-\left(z^2-y^2\right)\)
\(=\left(z^2-y^2\right)\left(81x^4-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(9x^2\right)^2-1^2\right]\)
\(=\left(z-y\right)\left(z+y\right)\left(9x^2-1\right)\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left[\left(3x\right)^2-1^2\right]\left(9x^2+1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(3x-1\right)\left(3x+1\right)\left(9x^2+1\right)\)
$---$
c, \(\dfrac{x^3}{8}-\dfrac{y^3}{27}+\dfrac{x}{2}-\dfrac{y}{3}\)
\(=\left[\left(\dfrac{x}{2}\right)^3-\left(\dfrac{y}{3}\right)^3\right]+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}\right)+\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\)
\(=\left(\dfrac{x}{2}-\dfrac{y}{3}\right)\left(\dfrac{x^2}{4}+\dfrac{xy}{6}+\dfrac{y^2}{9}+1\right)\)
$---$
d, \(x^6+x^4+x^2y^2+y^4-y^6\)
\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)+\left(x^4+x^2y^2+y^4\right)\)
\(=\left(x^4+x^2y^2+y^4\right)\left(x^2-y^2+1\right)\)
$Toru$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2+4y^2+4xy\)
\(=x^2+2.x.2y+\left(2y\right)^2\)
\(=\left(x+2y\right)^2\)
b) \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)\left(x+y+x-y\right)\)
\(=2y.2x\)
\(=4xy\)
c) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1-x-1\right)\left(3x+1+x-1\right)\)
a) \(x^6-y^6=\left(x^2\right)^3-\left(y^2\right)^3\)
\(=\left(x^2-y^2\right)\left(x^4+x^2y^2+y^4\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x2 - 2xy + y2 - 4m2 + 4mn - n2 = (x - y)2 - [(2m)2 - 2.2m.n + n2] = (x - y)2 - (2m - n)2
= [(x - y) - (2m - n)][(x - y) + (2m - n)] = (x - y - 2m + n)(x - y + 2m - n)
b) x2 - 4x2y2 + y2 + 2xy = x2 + 2xy + y2 - 4x2y2 = (x + y)2 - (2xy)2 = (x + y - 2xy)(x + y + 2xy)
c) x6 - y6 = (x3)2 - (y3)2 = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy + y2)(x + y)(x2 - xy - y2)
d) 25 - a2 + 2ab - b2 = 25 - (a2 - 2ab + b2) = 52 - (a - b)2 = (5 - a + b)(5 + a - b)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)
\(=\left(x^2-9\right)\left(x^2-1\right)+15\)
\(=x^4-10x^2+9+15\)
\(=x^4-10x^2+24\)
\(=\left(x^2-4\right)\left(x^2-6\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)
x6 + y6 = xy6
thiếu đề bài