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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)
c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
\(a,=\left(3x+1\right)^2-y^2=\left(3x-y+1\right)\left(3x+y+1\right)\\ b,=x\left(x^2-5x+6\right)=x\left(x^2-2x-3x+6\right)=x\left(x-2\right)\left(x-3\right)\)
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
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) \(x^2 (x+1)-2x(x+1)+x+1 \\ =(x+1)(x^2-2x+1)\\=(x+1)(x-1)^2\)
b) \(4x^2 -8x+3 \\= (2x^2)-2.2x .2 + 2^2 -1 \\=(2x-2)^2-1^2\\=(2x-2+1)(2x-2-1)\\= (2x-1)(2x-3)\)
\(a)x^5+x^4+1\)
\(=x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x+1\right)\)
\(b)x^8+x^7+1\)
\(=\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
\(#Tuyết\)
\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)
Xem lại đề ý b
a) \(x^6+1=x^6-\left(-1\right)=\left(x^3\right)^2-\left(-1^3\right)^2=\left(x^3\right)^2-\left(-1\right)\)
\(=\left(x^3-\left(-1\right)\right)\left(x^3+\left(-1\right)\right)=\left(x^3+1\right)\left(x^3-1\right)\)
b) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
c) \(x^9+1=\left(x^3\right)^3+\left(-1\right)^3\)
\(=\left(x^3+1\right)\left(\left(x^3\right)^2-x^3.1+1^2\right)=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
a) \(x^6+1=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
b) \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
c) \(x^9+1=\left(x^9-x^6+x^3\right)+\left(x^6-x^3+1\right)\)
\(=x^3\left(x^6-x^3+1\right)+\left(x^6-x^3+1\right)\)
\(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)