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a,\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)=3x^3y^3-x^2y^2+\frac{3}{5}x^3y^2\)
b,\(5x^3-5x=5x\left(x^2-1\right)=5x\left(x-1\right)\left(x+1\right)\)
c, \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(3x-5\right)\left(x-y\right)\)
1) 1/5x2y( 15xy2 - 5y + 3xy ) = 3x3y3 - x2y2 + 3/5x3y2
2) a) 5x3 - 5x = 5x( x2 - 1 ) = 5x( x2 - 12 ) = 5x( x - 1 )( x + 1 )
b) 3x2 + 5y - 3xy - 5x = ( 3x2 - 3xy ) + ( 5y - 5x )
= 3x( x - y ) + 5( y - x )
= 3x( x - y ) + 5[ -( x - y ) ]
= 3x( x - y ) - 5( x - y )
= ( 3x - 5 )( x - y )
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, \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)
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\(3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2\left(x-2y+3\right)\)
\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
\(12x^2y-18xy^2-3xy^2=3xy\left(4x-6y-y\right)\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(y\left(x-z\right)+7\left(z-x\right)=y\left(x-z\right)-7\left(x-z\right)=\left(x-z\right)\left(y-7\right)\)
\(27x^2\left(y-1\right)-9x^3\left(1-y\right)=27x^2\left(y-1\right)+9x^3\left(y-1\right)=9x^2\left(y-1\right)\left(3-x\right)\)
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
a)27x2.(y-1)9x3.(1-y)
=27x2.(y-1)+9x3.(y-1)
=9x2(y-1)[3+x]
b)8x3 + 1/27
=(2x)3 + (\(\frac{1}{3}\))3
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
= (2x+\(\frac{1}{3}\))(\(\left(2x\right)^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2\)
a) 27x2 ( y - 1) - 9x3 ( 1 - y)
=27x2 (y-1) + 9x3 ( y - 1 )
= (27x2 + 9x3) ( y -1 )
=9x2 ( x + 3) ( y - 1)
b)8x3+1/27
\(=\left(2x\right)^3+\left(\frac{1}{3}\right)^3\)
\(=\left(\frac{2x}{9}+\frac{1}{27}\right)\left(36x^2-6x+1\right)\)
c)49 ( y - 4 )2 - 9 ( y + 2)2
= [7(y - 4)]2 - [3(y + 2)]2
= (7y - 28 + 3y + 6)(7y - 28 - 3y - 6)
= (10y - 22)(4y - 34)
= 4(5y - 11)(2y - 34)
a, \(\left(25y^2+\dfrac{1}{9}x^2+\dfrac{5}{3}xy\right)\left(5y-\dfrac{1}{3}x\right)\)
Nó tự phân tích cho rồi nha bạn!
b, \(-125y^3-\dfrac{1}{27}x^3\)
\(=\left(-5y\right)^3-\left(\dfrac{1}{3}x\right)^3\)
\(=\left(-5y-\dfrac{1}{3}x\right)\left(25y^2-\dfrac{5}{3}xy+\dfrac{1}{9}x^2\right)\)
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Câu a :
\(\left(25y^2+\dfrac{1}{9}x^2+\dfrac{5}{3}xy\right)\left(5y-\dfrac{1}{3}x\right)\)
\(=\left(5y+\dfrac{1}{3}x\right)^2\left(5y-\dfrac{1}{3}x\right)\)
Câu b :
\(-125y^3-\dfrac{1}{27}x^3\)
\(=\left(-5y\right)^3-\left(\dfrac{1}{3}x\right)^3\)
\(=\left(-5y-\dfrac{1}{3}x\right)\left(25y^2+\dfrac{5}{3}xy+\dfrac{1}{9}x^2\right)\)