Bài 2: đưa về dạng lập phương của 1 tổng, 1 hiệu.
1, x3-9x2y+27xy2-27y3
2, 27x3-9x2y+xy2-1/27y3
3, x6-3x4y+3xy2-y3
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\(1,\\ a,=\left(x+2\right)\left(x^2-2x+4\right)\\ b,=\left(x-4\right)\left(x^2+8x+16\right)\\ c,=\left(3x+1\right)\left(9x^2-3x+1\right)\\ d,=\left(4m-3\right)\left(16m^2+12m+9\right)\\ 2,\\ a,=x^3+125\\ b,=1-x^3\\ c,=y^3+27t^3\)
a)
\(=\left(x+2\right)\left(x^2-2x+4\right)\)
b)
\(=\left(x-4\right)\left(x^2+4x+16\right)\)
c)=\(\left(3x+1\right)\left(9x^2-3x+1\right)\)
d)
=\(\left(4m-3\right)\left(16m^2+12m+9\right)\)
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)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
Ta có: P(x) + Q(x)
= (3x2y - 2x + 5xy2 - 7y2 ) + (3xy2 - 7y2 - 9x2y - x - 5)
= -6x2y + 8xy2 - 14y2 - 3x - 5. Chọn A
\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)
M = P + Q
= (3x2y − 2x + 5xy2 − 7y2) + (3xy2 − 7y2 − 9x2y – x – 5)
= 3x2y − 2x + 5xy2 − 7y2 + 3xy2 − 7y2 − 9x2y – x – 5
= (5xy2 + 3xy2) + (3x2y – 9x2y) – (2x + x) – (7y2 + 7y2) – 5
= 8xy2 − 6x2y − 3x − 14y2 – 5.
M = Q – P
= (3xy2 − 7y2 − 9x2y – x – 5) - (3x2y − 2x + 5xy2 − 7y2)
= 3xy2 – 7y2 – 9x2y – x – 5 – 3x2y + 2x – 5xy2 + 7y2.
= (3xy2 – 5xy2) – (9x2y + 3x2y) + (2x – x) + (-7y2 + 7y2) – 5
= -2xy2 − 12x2y + x – 5
\(x^3+12x^2+48x+64=x^3+3.x^2.4+3.x.4^2+4^3=\left(x+4\right)^3\)
\(x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3=\left(x-2\right)^3\)
\(a,x^3+6x^2y+12xy^2+8y^3\\ =x^3+3.2x^2+3.2^2.x+\left(2y\right)^3\\ =\left(x+2y\right)^3\)
\(b,x^3-3x^2+3x-1\\ =x^3-3x^2.1+3x.1^2-1^3\\ =\left(x-1\right)^3\)
a) \(x^3+6x^2y+12xy^2+8y^3\)
\(=x^3+3\cdot x^2\cdot2y+2\cdot x\cdot\left(2y\right)^2+\left(2y\right)^3\)
\(=\left(x+2y\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
1, x3-9x2y+27xy2-27y3=(x-3y)3
2, 27x3-9x2y+xy2-\(\dfrac{1}{27}\)y3=(3x-\(\dfrac{1}{3}\)y)3
3)x6-3x4y+3xy2-y3=(x2-y)3
1) \(x^3-9x^2y+27xy^2-27y^3=\left(x-3y\right)^3\)
2) \(27x^3-9x^2y+xy^2-\dfrac{1}{27}y^3=\left(3x-\dfrac{1}{3}y\right)^3\)
3) \(x^6-3x^4y+3xy^2-y^3=\left(x^2-y\right)^3\)