Phân tích nhân tử bằng hằng đẳng thức.............tìm x?
a)2-25x^2=0
b)x^6-y^6
c) x^16-1
Update:d) x^3-6x^2+12x-8=0
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a)2-25x^2=0
b)x^6-y^6
c) x^16-1
Update:d) x^3-6x^2+12x-8=0
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Dễ thôi em ạ !!!!!!!
a) 2-25x^2 = 0
<=> √2^2 - (5x)^2 = 0
<=> (√2+5x)(√2-5x)=0
<=>√2+5x=0 hoặc √2-5x=0
suy ra x = -√2/5 hoặc x = √2/5
b)x^6-y^6
= (x^3)^2 - (y^3)^2
= (x^3-y^3)(x^3+y^3)
=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)
c) x^16-1
=(x^8)^2-1
=(x^8+1)(x^8-1)
d) x^3-6x^2+12x-8=0
<=> x^3-3.x^2.2+3.x.2^2-2^3=0
<=> (x-2)^3=0
<=> x-2=0
suy ra x=2.
Nhớ cho chị 5* nhé !!!!!!!
a) 2-25x^2 = 0
<=> â2^2 - (5x)^2 = 0
<=> (â2+5x)(â2-5x)=0
<=>â2+5x=0 hoặc â2-5x=0
suy ra x = -â2/5 hoặc x = â2/5
b)x^6-y^6
= (x^3)^2 - (y^3)^2
= (x^3-y^3)(x^3+y^3)
=(x-y)(x^2+xy+y^2)(x+y)(x^2-xy+y^2)
c) x^16-1
=(x^8)^2-1
=(x^8+1)(x^8-1)
d) x^3-6x^2+12x-8=0
<=> x^3-3.x^2.2+3.x.2^2-2^3=0
<=> (x-2)^3=0
<=> x-2=0
suy ra x=2.
a/(cÄn 2-5x)(cÄn 2+5x)
b/(x^3-y^3)(x^3+y^3)
=(x+y)^2(x^2+xy+y^2)(x^2-xy+y^2)
c/ (x^8-1)(x^8+1)
=(x^4+1)(x^4-1)...
=(x+1)(x-1)(x^2+1)(x^4+1)(x^8+1)
d/ x^3-3*2x^2+3.2^2x-2^3=(x-2)^3.