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1/3x + y =-1 => x +3y = -3
\(A=x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3=\left(-3\right)^3=-27\)
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: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
a) \(-x^3+9x^2-27x+27=-\left(x^3-3.3.x^2+3.3^2.x-3^3\right)=-\left(x-3\right)^3\)
b)\(x^4-2x^3-x^2+2x+1=x^4+\left(-x\right)^2+\left(-1\right)^2+2x^2\left(-x\right)+2.\left(-x\right).\left(-1\right)+2x^2.\left(-1\right)\)
\(=\left(x^2-x-1\right)^2\)
c)\(8x^3+27y^3+36x^2y+54xy^2=\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(2x+3y\right)^2\)
a) x^3-2x^2+x
= x(x^2-2x+1)
x(x-1)^2
b) x^2-2x-15
= (x^2-2x+1)-16
= (x-1)^2-4^2
= ( x-5)(x+3)
d) 5(x-y)-y(x-y)
=(x-y)(5-y)
e) 27x^2(y-1)-9x^3(1-y)
= 27x^2(y-1)+9x^3(y-1)
= (y-1)(27x^2+9x^3)
= 9x^2(y-1)(3+x)
f) 36-12x+x^2
= x^2- 6.2x+36
= ( x - 6)^2
g) 125^3+27y^3
125^3+ ( 9y)^3
= 125^3+ 3.125^2.9y+3.125.(9y)2+ (9y)3
i) xy+xz+3y+3z
= x( x+y) + 3(y+z)
= (y+z)(x+3)
a/ \(x^3-2x^2+x\)
= \(x\left(x^2-2x+1\right)\)
= \(x\left(x-1\right)^2\)
b/ \(x^2-2x-15\)
= \(x^2-2x+1-16\)
= \(\left(x-1\right)^2-4^2\)
= \(\left(x-1-4\right)\left(x-1+4\right)\)
= \(\left(x-5\right)\left(x+3\right)\)
c/ \(x^2-2x-y^2+1\)
= \(\left(x-1\right)^2-y^2\)
= \(\left(x-y-1\right)\left(x+y-1\right)\)
d/ \(5\left(x-y\right)-y\left(x-y\right)\)
= \(\left(x-y\right)\left(5-y\right)\)
e/ \(27x^2\left(y-1\right)-9x^3\left(1-y\right)\)
= \(27x^2\left(y-1\right)+9x^3\left(y-1\right)\)
= \(\left(y-1\right)\left(27x^2+9x^3\right)\)
f/ \(36-12x+x^2\)
= \(x^2-12x+6^2\)
= \(\left(x-6\right)^2\)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
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) \(x^6-3x^4y+3xy^2-y^3=\left(x^2-y\right)^3\)