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\(a,3x-15xy=3x\left(1-5y\right)\\ ---\\ 8x^2+6x-4=2\left(4x^2+3x-2\right)\\ ---\\ 5x^2+25xy+10y^2=5\left(x^2+5xy+2y^2\right)\\ ---\\ 9x^2y^2+6x^2y-\dfrac{1}{2}xy^2=\dfrac{1}{2}xy\left(18xy+12x-y\right)\)
a)15x^2y-20xy^2-25xy=5xy(3x^2-4y-5)
b) x^2-9-2xy+y^2=(x^2-2xy+y^2)-9=(x-y)^2-3^2=(x-y-3)(x-y+3)
c)2x^2-5x-7=2x^2 -7x+2x-7 = x(2x-7)+(2x-7) =(2x-7)(x+1)
=(x^5-5x^2)-(2xy+10y)
=(x^5-5x^2)-(2xy-10y)
=x^2.(x-5)-2y.(x-5)
=(x^2-2y).(x-5)
k nha
\(x^3-5x^2-2xy+10y=x^2\left(x-5\right)-2y\left(x-5\right)=\left(x-5\right)\left(x^2-2y\right)=\left(x-5\right)\left(x-2y\right)\left(x+2y\right)\)
\(x^2-5x+2xy-10y\)
\(=\left(x^2-5x\right)+\left(2xy-10y\right)\)
\(=x\left(x-5\right)+2y\left(x-5\right)\)
\(=\left(x+2y\right)\left(x-5\right)\)
\(=x^2y\left(x-5\right)-2y\left(x-5\right)+0\)
\(=\left(x-5\right)\left(x^2y-2y\right)\)
xong phân tích nốt cái bậc 2 kia để được max điểm :)))
1) 4x2 + 5x - 6 = 4x2 + 8x - 3x - 6 = 4x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( 4x - 3 )
2) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
3) 2x2 + 3x - 27 = 2x2 - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 ) < đã sửa ._. >
4) 7x2 + 3xy - 10y2 = 7x2 - 7xy + 10xy - 10y2 = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )
5) x2 + 5x - 2 < sai đề ._. >
6) x8 + x7 + 1 = x8 + x7 + x6 - x6 + 1
= ( x8 + x7 + x6 ) - ( x6 - 1 )
= x6( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x6( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )[ x6 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )
a, x2-5xy+2x-10y = (x2 + 2x)-(5xy+10y)
= x(x+2)-5y(x+2)
= (x+2)(x-5y)
b, x2-5x+4 = x2- x - 4x +4
= (x2-x)-(4x-4)
=x(x-1)-4(x-4)
=(x-1)(x-4)
\(a,x^2-5xy+2x-10y\)
\(=\left(x^2-5xy\right)+\left(2x-10y\right)\)
\(=x\left(x-5y\right)+2\left(x-5y\right)\)
\(=\left(x-5y\right)\left(x+2\right)\)
\(b,x^2-5x+4\)
\(=x^2-4x-x+4\)
\(=x\left(x-4\right)-\left(x-4\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
\(=2xy+5x+4y^2+10y\)
\(=x\left(2y+5\right)+2y\left(2y+5\right)\)
\(=\left(x+2y\right)\left(2y+5\right)\)
\(=2x\left(x+2y\right)+5\left(x+2y\right)=\left(x+2y\right)\left(2x+5\right)\)
\(-5x^2-25xy+10y^2=-5\left(x^2+5xy-2y^2\right)\)