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\(1,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(3,7x-6x^2-2=-6x^2+7x-2=-6x^2+3x+4x-2=3x\left(-2x+1\right)+2\left(2x-1\right)\)
\(=3x\left(1-2x\right)-2\left(1-2x\right)=\left(1-2x\right)\left(3x-2\right)\)
\(2,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=3x-6x^2+4x-2=3x\left(1-2x\right)-2\left(1-2x\right)=\left(3x-2\right)\left(1-2x\right)\)
\(a,x^2+5x-6=x^2-x+6x-6\)
\(=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(b,5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
\(c,7x-6x^2-2=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)=\left(2x-1\right)\left(2-3x\right)\)
Làm 1 cách là đủ rồi mà (: 6 cách thì đến bao giờ :v
a) x2 + x - 6 = x2 - 2x + 3x - 6 = x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( x + 3 )
b) x2 - 4x + 3 = x2 - x - 3x + 3 = x( x - 1 ) - 3( x - 1 ) = ( x - 1 )( x - 3 )
c) x2 + 5x + 4 = x2 + x + 4x + 4 = x( x + 1 ) + 4( x + 1 ) = ( x + 1 )( x + 4 )
d) x2 - x - 6 = x2 + 2x - 3x - 6 = x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x - 3 )
e) 2x2 + 5x + 3 = 2x2 + 2x + 3x + 3 = 2x( x + 1 ) + 3( x + 1 ) = ( x + 1 )( 2x + 3 )
g) 2x2 - 7x + 3 = 2x2 - 6x - x + 3 = 2x( x - 3 ) - ( x - 3 ) = ( x - 3 )( 2x - 1 )
h) 3x2 + 10x - 8 = 3x2 + 12x - 2x - 8 = 3x( x + 4 ) - 2( x + 4 ) = ( x + 4 )( 3x - 2 )
k) \(\frac{1}{2}x^2-\frac{19}{6}x+1=\frac{1}{2}x^2-\frac{1}{6}x-3x+1=\frac{1}{2}x\left(x-\frac{1}{3}\right)-3\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{3}\right)\left(\frac{1}{2}x-3\right)\)
\(x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x+6\right)\left(x-1\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-\left(6x^2-7x+2\right)=-\left[\left(6x^2-3x\right)-\left(4x+2\right)\right]=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left[\left(3x-2\right)\left(2x-1\right)\right]\)
d) \(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+3\right)\left(x+1\right)\)
a) đề sai thì phải
b) 2x2 - 2xy - 7x + 7y = (2x2 - 2xy) - (7x - 7y) = 2x(x - y) - 7(x - y) = (x - y)(2x - 7)
c) x2 - 3x + xy - 3y = (x2 + xy) - (3x + 3y) = x(x + y) - 3(x + y) = (x + y)(x - 3)
đ) x2 - xy + x - y = (x2 - xy) + (x - y) = x(x - y) + (x - y) = (x - y)(x + 1)
Phối hợp các phương pháp
a) x2 - 2xy + y2 - xy + y2 = (x2 - 2xy + y2) - (xy - y2) = (x - y)2 - y(x - y) = (x - y)(x - y - y) = (x - y)(x - 2y)
\(x^2+5x-6\\ =x^2-x+6x-6\\ =x\left(x-1\right)+6\left(x-1\right)\\ =\left(x+6\right)\left(x-1\right)\\ ---\\ 5x^2+5xy-x-y\\ =x\left(5x-1\right)+y\left(5x-1\right)\\ =\left(5x-1\right)\left(x+y\right)\\ ----\\ 7x-6x^2-2\\ =-6x^2+3x+4x-2\\ =-3x\left(2x-1\right)+2\left(2x-1\right)\\ =\left(2-3x\right)\left(2x-1\right)\)
\(x^2+5x-6=\left(x-1\right)\left(x+6\right)\\ 5x^2+5xy-x-y=\left(5x-1\right)\left(x+y\right)\\ 7x-6x^2-2=\left(3x-2\right)\left(2x-1\right)\)
Ta có : x2 - x + 6x - 6
= x(x - 1) + 6(x - 1)
= (x + 6)(x - 1)
b) 5x2 + 5xy - x - y
= 5x(x + y) - (x + y)
= (5x - 1)(x + y)
câu a,b làm như trên. câu c:
c, 7x - 6x2 - 2 = -6x^2 + 7x - 2 = -( 6x^2-3x-4x+2) = -[ 3x ( 2x-1)-2(2x-1)]=-(3x-2)(2x-1)
a) \(x^2+5x-6=x^2-x+6x-6\)
=\(\left(x^2-x\right)+\left(6x-6\right)\)
\(=x\left(x-1\right)+6\left(x-1\right)\)
\(=\left(x-1\right)\left(x+6\right)\)
x^2 + 5x -6
= x^2 + 5x - (5+1)
= x^2 + 5x -5 -1
= 5(x-1) + (x^2 -1)
= 5(x-1) + (x-1) (x+1)
= (5+x+1) (x-1)
5x^2 + 5xy -x-y
= 5x(x+y) - (x+y)
= (5x -1) (x+y)
7x - 6x^2 - 1 (câu này tớ tự ý sửa đề chút ^^!)
= 6x + x - 6x^2 -1
= 6x (1-x) - (1-x)
= (6x -1) (1-x)
x2 + 5x - 6= x2 - x + 6x -6= x(x-1) + 6(x-1)= (x+6)(x-1)
5x2 + 5xy - x - y= 5x2 - x + 5xy - y= x(5x-1) + y(5x - 1)= ( x +y)( 5x -1)
7x - 6x2 - 2= -6x2 + 3x + 4x -2= -3x(2x - 1) + 2(2x -1)= (2 - 3x)(2x-1)