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a)
\(2x^2y-8xy^2\\ =2xy\left(x-4y\right)\)
b)
\(x^2-2xy+y^2-16\\ =\left(x^2-2xy+y^2\right)-16\\ =\left(x-y\right)^2-16\\ =\left(x-y-4\right)\left(x-y+4\right)\)
a) \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
b) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
a) = x2 + 2x + 3x + 6 = x( x + 2 ) + 3( x + 2 ) = ( x + 2 )( x + 3 )
b) = x2 - x + 4x - 4 = x( x - 1 ) + 4( x - 1 ) = ( x - 1 )( x + 4 )
c) = ( x2 - 10x + 25 ) - 9 = ( x - 5 )2 - 32 = ( x - 8 )( x - 2 )
d) = 6x2 - 15x + 8x - 20 = 3x( 2x - 5 ) + 4( 2x - 5 ) = ( 2x - 5 )( 3x + 4 )
a) x2 + 5x + 6 = x2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2) = (x + 3)(x + 2)
b) x2 + 3x - 4 = x2 - x + 4x - 4 = x(x - 1) + 4(x - 1) = (x + 4)(x - 1)
c) x2 - 10x + 16 = x2 - 2x - 8x + 16 = x(x - 2) - 8(x - 2) = (x - 8)(x - 2)
d) 6x2 - 7x - 20 = 6x2 + 8x - 15x - 20 = 2x(3x + 4) - 5(3x + 4) = (2x - 5)(3x + 4)
a, 2x2 + 10xy=2x(x+5y)
b, 3x ( y - x ) + 6y ( y - x )=(3x+6y)(y-x)
c, 3x ( x - 2 ) - x + 2 + 5x ( x - 2 )=3x(x-2)-(x-2)+5x(x-2)=(8x-1)(x-2)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a) 5x2 -20
= 5(x2 -4)
=5 (x2 -22)
= 5(x-2)(x+2)
b) 16 - (x+y)2
=42 -(x+y)2
= (4-x-y)(4+x+y)
a, \(5\left(x^2-4\right)=5\left(x-2\right)\left(x+2\right)\)
b, \(16-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)
mấy bài này áp dụng hđt là được nhé