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a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
a. x3 - 3x2 + 3x - 1
= (x-1)3
b. (x+y)2 - 4x2
=(x+y-4x)(x+y+4x)
c. 27x3 + 1/8
= (3x)3 +(1/2)3
=(3x+ 1/2) (9x - 3x.1/2 - 1/4)
d. ( x+y)3 - (x-y)3
= [(x+y)-(x-y)] [(x+y)2 + (x+y)(x-y) + (x-y)2]
=(x+y-x+y)[x2+2xy+y2+x2-y2+x2-2xy+y2)
=2y . (3x2+y2)
Mấy câu này ko biết đúng hay sai :{
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
a ) 36x2 - ( 3x - 2 )2
= ( 6x - 3x + 2 ) ( 6x + 3x - 2 )
= ( 3x + 2 ) ( 9x - 2 )
b ) 16.( 4x + 5 )2 - 25. ( 2x + 2 )2
= [ 4.( 4x + 5 ) + 5. ( 2x + 2 ) ] [ 4 .( 4x + 5 ) - 5. ( 2x + 2 ) ]
= ( 16x + 5 + 10x + 10 ) ( 16x + 5 - 10x - 10 )
= ( 26x + 15 ) ( 6x - 5 )
Lời giải:
a) $x^3+3x^2y+x+3xy^2+y+y^3$
$=(x^3+3x^2y+3xy^2+y^3)+(x+y)$
$=(x+y)^3+(x+y)=(x+y)[(x+y)^2+1]$
b) $x^3+y(1-3x^2)+x(3y^2-1)-y^3$
$=(x^3-3x^2y+3xy^2-y^3)-(x-y)$
$=(x-y)^3-(x-y)=(x-y)[(x-y)^2-1]=(x-y)(x-y-1)(x-y+1)$
c)
$27x^3+27x^2+9x+1=(3x+1)^3$
d)
$x(x+1)^2+x(x-5)-5(x+1)^2$
$=x(x+1)^2-5(x+1)^2+x(x-5)$
$=(x-5)(x+1)^2+x(x-5)=(x-5)[(x+1)^2+x]$
$=(x-5)(x^2+3x+2)=(x-5)(x+1)(x+2)$
a) x3+ 6x2+12x+8
=(x+2)3
b)x3-3x2+3x-1
=(x-1)3
c)1-9x+27x2-27x3
=(1-3x)3
d)x3 +\(\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)
=(x+\(\frac{1}{2}\))3 ( phần này mik là khác đầu bài bạn đi 1 chút nhưng mik tôn trọng ý kiến của bạn hơn nên mik nghĩ mik làm sai)
e) 27x3-54x2y+36xy2-8y3
=(3x-2y)2
a) x3 + 6x2 + 12x + 8
= (x^3+2^3)+6x.(x+2)
= (x+2).(x^2-2x+4)+6x(x+2)
= (x+2).(x^2+4x+4)
b) x3 - 3x2 + 3x - 1
= (x^3-1) -3x.(x-1)
= (x-1).(x^2+x+1) - 3x(x-1)
= (x-1).(x^2-2x+1)
Câu d ko hiểu đề :v
e) 27x3- 54 x2y + 36 xy2 - 8y3
= (27x^3-8y^3)-(54x^2y+36xy^2)
= (3x-2y).(9x^2+6xy+4y^2)-18xy(3x-2y)
= (3x-2y).(9x^2-12xy+4y^2)
Thế nhé :)
a) 1 - 2y + y2
= (1-y)2
b) ( x + 1 )2 - 25
=( x + 1 )2 - 52
=(x+1+5)(x+1-5)
\(x^3+\frac{1}{x^3}=x^3+\left(\frac{1}{x}\right)^3=\left(x+\frac{1}{x}\right)\left(x^2-x+\frac{1}{x^2}\right)\)( x khác 0 )
\(-x^3+9x^2-27x+27=-\left(x^3-9x^2+27x-27\right)=-\left(x-3\right)^3\)
\(\left(xy+1\right)^2-\left(x-y\right)^2=\left(xy+1-x+y\right)\left(xy+1+x-y\right)\)
\(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)
b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)
c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
a) Ta có: \(x^2+2x+1\)
\(=x^2+2\cdot x\cdot1+1^2\)
\(=\left(x+1\right)^2\)
b) Ta có: \(1-2y+y^2\)
\(=y^2-2\cdot y\cdot1+1^2\)
\(=\left(y-1\right)^2\)
c) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
d) Ta có: \(27+27x+9x^2+x^3\)
\(=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+9\right)\)
\(=\left(x+3\right)^3\)
e) Ta có: \(8-125x^3\)
\(=2^3-\left(5x\right)^3\)
\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) Ta có: \(64x^3+\frac{1}{8}\)
\(=\left(4x\right)^3+\left(\frac{1}{2}\right)^3\)
\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
g) Ta có: \(1-x^2y^4\)
\(=1^2-\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)\left(1+xy^2\right)\)
a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)
b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)
c) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)
e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
Ko chắc ạ!
a, \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)
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