phân tích đa thức thành nhân tu
a/ 9x^2 -12x^2y^3+16y^6 b/ 4x^4 -16x^2y^2+16y^6
c/8x^3+27y^3 d/a^3+1/27
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a) 9x4+16y6-24x2y3
=(3x2)2-2.3x2.4y3+(4y3)2
=(3x2-4y3)2
b) 16x2-24xy+9y2
=(4x)2-2.4x.3y+(3y)2
=(4x-3y)2
c) 36x2-(3x-2)2
=(36x-3x+2)(36x+3x-2)
=(33x+2)(39x-2)
d) 27x3+54x2y+36xy2+8y3
=(3x)3+3.(3x)2.2y+3.3x.(2y)2+(2y)3
=(3x+2y)3
e) y9-9x2y6+27x4y3-27x6
=(y3)3-3.(y3)2.3x2+3.y3.(3x2)2-(3x2)3
=(y3-3x2)3
f) 64x3+1
= (4x)3+13
=(4x+1)[(4x)2-4x.1+12]
=(4x+1)(16x2-4x+1)
e) 27x6-8x3 *sửa đề*
=(3x2)3-(2x)3
=(3x2-2x)[(3x)2+3x2.2x+(2x)2]
=(3x2-2x)(9x2+6x3+4x2)
~~~
a) \(4x^2-12x+9=\left(2x\right)^2-2\cdot2x\cdot3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2\cdot2x\cdot1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2\cdot1\cdot6x+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2\cdot3x\cdot4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
e) \(8x^3+1=\left(2x\right)^3+1^3=\left(2x+1\right)\left(4x^2+2x+1\right)\)
f) \(-8x^3+27=3^3-\left(2x\right)^3=\left(3-2x\right)\left(9+6x+4x^2\right)\)
a)x2-4y2
=x2-(2y)2
=(x-2y)(x+2y)
b)x3+27y3
=x3+(3y)3
=(x+3y)(x2-3xy+9y2)
c)4x2+12xy+9y2-16
=(2x+3y)2-42
=(2x+3y-4)(2x+3y+4)
d)9x2-24xy+16y2-64
=(3x-4y)2-82
=(3x-4y-8)(3x-4y+8)
e)8x3-27y3
=(2x)3-(3y)3
=(2x-3y)(4x2+6xy+9y2)
f)5x3-7x2+10x-14
=5x3+10x-7x2-14
=5x(x2+2)-7(x2+2)
=(5x-7)(x2+2)
\(9x^4-12x^2y^3+16y^6=\left(3x^2+4y^3\right)^2\)
\(4x^4-16x^2y^3+16y^6=\left(2x^2-4y^3\right)^2\)
\(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
\(a^3+\frac{1}{27}=\left(a+\frac{1}{3}\right)\left(a^2-\frac{1}{3}a+\frac{1}{9}\right)\)
phân a bn phân tích kiểu j hay thế :D