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3 tháng 8 2019

a, \(m^3+27\)

\(\Leftrightarrow m^3+3^3\)

\(\Leftrightarrow\left(m+3\right)\left(m^2-m.3+3^2\right)\)

\(\Leftrightarrow\left(m+3\right)\left(m^2-3m+9\right)\)

b,\(\frac{1}{27}+a^3\)

\(\Leftrightarrow\frac{1}{27}\left(1+27a^3\right)\)

\(\Leftrightarrow\frac{1}{27}.\left(1+3a\right)\left(1-3a+9a^2\right)\)

c,\(\left(a+b\right)^3-c^3\)

\(\Leftrightarrow\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]\)

\(\Leftrightarrow\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)

d,\(x^9+1\)

\(\Leftrightarrow\left(x^3+1\right)\left(x^6-x^3+1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)

e,\(x^3+9x^2+27x+27\)

\(\Leftrightarrow x^3+3.x^2.3+3x.9+3^3\)

\(\Leftrightarrow x^3+3x^2.3+3x+3^2+3^3\)

\(\Leftrightarrow\left(x+3\right)^3\)

AH
Akai Haruma
Giáo viên
23 tháng 6 2024

Lời giải:

$x^3-4x^2-12x+27$

$=(x^3+3x^2)-(7x^2+21x)+(9x+27)$

$=x^2(x+3)-7x(x+3)+9(x+3)$

$=(x+3)(x^2-7x+9)$

 

29 tháng 6 2017

Ta có : x+ 2x2 + 2x + 1 

= x3 + x2 + (x2 + 2x + 1)

= x2(x + 1) + (x + 1)2

= (x + 1) ( x+ x + 1)

29 tháng 6 2017

a)\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2\)

                                                       \(=a^2b^2+a^2+b^2+1\)

                                                        \(=a^2\left(b^2+1\right)+\left(b^2+1\right)\)

                                                       \(\left(a^2+1\right)\left(b^2+1\right)\)

b)\(x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1\)

                                        \(=x^2\left(x+1\right)+x\left(x+1\right)+x+1\)

                                          \(=\left(x^2+x+1\right)\left(x+1\right)\)

                                           

29 tháng 11 2018

\(x^3-4x^2+12x-27\)

\(=x^3-3x^2-x^2+3x+9x-27\)

\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-x+9\right)\)

28 tháng 1 2020

i. (x2+y2+z2).(x+y+z)2+(xy+yz+zx)2

1 tháng 11 2017

\(a.\)

\(\left(x-9\right)^2+12x\left(x-3\right)^2\)

\(\Rightarrow\left(x-3\right)\left(x+3\right)+12x\left(x-3\right)^2\)

\(\Rightarrow\left(x-3\right)\left(x+3+12x+x-3\right)\)

\(\Rightarrow14x\left(x-3\right)\)

\(b.\)

\(a\left(b^2+c^2\right)-b\left(c^2+a^2\right)+c\left(a^2+b^2\right)-2abc\)

\(=ab^2+ac^2-bc^2-ba^2+\left(ca^2+cb^2-2abc\right)\)

\(=ab\left(b-a\right)+c^2\left(a-b\right)+c\left(a-b\right)^2\)

\(=c^2\left(a-b\right)-ab\left(a-b\right)+c\left(a-b\right)^2\)

\(=\left(a-b\right)\left(c^2-ab+ac-bc\right)\)

\(=\left(a-b\right)\left[c\left(c+a\right)-b\left(c+a\right)\right]\)

\(=\left(a-b\right)\left(c-b\right)\left(c+a\right)\)

\(c.\)

\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=a^3+b^3+3ab\left(a+b\right)+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

1 tháng 11 2017

a) \(\left(x^2-9\right)^2+12x\left(x-3\right)^2\)

\(=\left[\left(x-3\right)\left(x+3\right)\right]^2+12x\left(x-3\right)^2\)

\(=\left(x-3\right)^2\left(x+3\right)^2+12x\left(x-3\right)^2\)

\(=\left(x-3\right)^2\left[\left(x+3\right)^2+12x\right]\)

\(=\left(x-3\right)^2\left(x^2+6x+3^2+12x\right)\)

\(=\left(x-3\right)^2\left(x^2+18x+9\right)\)

19 tháng 8 2016

Bài 1 :

\(x^2+4x-y^2+4\)

\(=\left(x^2+4x+4\right)-y^2\)

\(=\left(x+2\right)^2-y^2\)

\(=\left(x+2+y\right)\left(x+2-y\right)\)

Bài 2 : Ta có : \(a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=-c^3\)

\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3-3abc=-c^3\) ( Vì \(a+b=-c\) )

\(\Rightarrow a^3+b^3+c^3=3abc\)

19 tháng 8 2016

Bài 1:

x2 +4x-y2+4

=(x2+4x+4)-y2

=(x+2)2-y2

=(x-y+2)(x+y+2)

Bài 2:

 a3+b3+c3 =  3abc

=>a3+b3+c3-3abc=0

=>[(a+b)3+c3]-3ab(a+b)-3abc=0

=>(a+b+c)[(a+b)2-(a+b)c+c2]-3ab(a+b+c)=0

=>(a+b+c)(a2+b2+c2-ac-bc-ab)=0

Từ a+b+c=0

=>0*(a2+b2+c2-ac-bc-ab)=0 (luôn đúng)