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

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

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, \(x^6+x^4+x^2y^2+y^4-y^6\)

\(=\left(x^6-y^6\right)+\left(x^4+x^2y^2+y^4\right)\)

\(=\left[\left(x^2\right)^3-\left(y^2\right)^3\right]+\left(x^4+x^2y^2+y^4\right)\)

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

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

$Toru$

a) (x - y)(x + y + 3).                    b) (x + y - 2xy)(2 + y + 2xy).

c) x 2 (x + l)( x 3  -  x 2  + 2).              d) (x – 1 - y)[ ( x   -   1 ) 2   +   ( x   -   1 ) y   +   y 2 ].

ủa mũ hay để nguyên z?? 

hđt =x^2(x+1)^2(x^2-2x+2)

x⁶ - x⁴ + 2x³ + 2x²

= x²(x⁴ - x² + 2x + 2)

= x²[(x⁴ - x²) + (2x + 2)]

= x²[x²(x² - 1) +2(x + 1)]

= x²[x²(x - 1)(x + 1) + 2(x + 1)]

= x²(x + 1)[x²(x - 1) + 2]

= x²(x + 1)(x³ - x² + 2)

= x²(x + 1)(x³ + x² - 2x² - 2x + 2x + 2)

= x²(x + 1)[(x³ + x²) - (2x² + 2x) + (2x + 2)]

= x²(x + 1)[x²(x + 1) - 2x(x + 1) + 2(x + 1)]

= x²(x + 1)²(x² - 2x + 2)

a,

\(A=4(x-2)(x+1)+(2x-4)^2+(x+1)^2\\=[2(x-2)]^2+2\cdot2(x-2)(x+1)+(x+1)^2\\=[2(x-2)+(x+1)]^2\\=(2x-4+x+1)^2\\=(3x-3)^2\)

Thay $x=\dfrac12$ vào $A$, ta được:

\(A=\Bigg(3\cdot\dfrac12-3\Bigg)^2=\Bigg(\dfrac{-3}{2}\Bigg)^2=\dfrac94\)

Vậy $A=\dfrac94$ khi $x=\dfrac12$.

b,

\(B=x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\\=(x^9-1)-(x^7-x^4)-(x^6-x^3)-(x^5-x^2)\\=[(x^3)^3-1]-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1)-x^4(x^3-1)-x^3(x^3-1)-x^2(x^3-1)\\=(x^3-1)(x^6+x^3+1-x^4-x^3-x^2)\\=(x^3-1)(x^6-x^4-x^2+1)\)

Thay $x=1$ vào $B$, ta được:

\(B=(1^3-1)(1^6-1^4-1^2+1)=0\)

Vậy $B=0$ khi $x=1$.

$Toru$

a) x6 – x4 + 2x3 + 2x2

= x2(x4 – x2 + 2x + 2)

= x2[x2(x2 – 1) + 2(x + 1)]

= x2. [x2.(x -1).(x + 1) + 2(x+ 1)]

= x2 (x+ 1).[x2(x- 1)+ 2]

= x2(x + 1)(x3 – x2 + 2)

= x2(x + 1)[(x3 + 1) – (x2 – 1)]

= x2(x + 1).[(x + 1).(x2 – x + 1) - (x - 1).(x + 1)]

= x2(x + 1)(x + 1)( x2 – x + 1 – x + 1)

= x2(x + 1)2(x2 – 2x + 2).

b) 4x4 + y4 = 4x4 + 4x2y2 + y4 - 4x2y2

= (2x2 + y2)2 - (2xy)2

= (2x2 + y2 + 2xy)(2x2 + y2 - 2xy)

Tròn đã làm bằng cách:

\(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3\)

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

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

\({x^6} + {y^6} = {\left( {{x^2}} \right)^3} + {\left( {{y^2}} \right)^3} = \left( {{x^2} + {y^2}} \right)\left[ {{{\left( {{x^2}} \right)}^2} - {x^2}.{y^2} + {{\left( {{y^2}} \right)}^2}} \right] = \left( {{x^2} + {y^2}} \right)\left( {{x^4} - {x^2}{y^2} + {y^4}} \right)\)

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)

xin lỗi các bạn, đề mink có vấn đề: ý c phải là: x^6 - y^6

\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)=\)\(\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)