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Sửa đề : \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(9x+9y-9\right)^2-\left(8x+12y+4\right)^2\)
\(=\left(9x+9y-9-8x-12y-4\right)\left(9x+9y-9+8x+12y+4\right)\)
\(=\left(x-3y-13\right)\left(17x+21y-5\right)\)
Đúng là Tú có khác (:
9( x + y - 1 )2 - 4( 2x + 3y + 1 )2
= 32( x + y - 1 )2 - 22( 2x + 3y + 1 )2
= [ 3( x + y - 3 ) ]2 - [ 2( 2x + 3y + 1 ) ]2
= ( 3x + 3y - 3 )2 - ( 4x + 6y + 2 )2
= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]
= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )
= ( -x - 3y - 5 )( 7x + 9y - 1 )
\(x^4-y^2\left(2x-y\right)^2\)
\(=x^4-\left(2xy-y^2\right)^2\)
\(=\left(x^2-2xy+y^2\right)\left(x^2+2xy-y^2\right)\)
\(=\left(x-y\right)^2\left(x^2+2xy-y^2\right)\)
a: 2x+4=2(x+2)
b: \(x^2+2xy+y^2-9=\left(x+y-3\right)\left(x+y+3\right)\)
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
Bài làm:
Ta có: \(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
Học tốt!!!!
Ta có :
\(9\left(x-y\right)^2-4\left(x+y\right)^2=9x^2-18xy+9y^2-4x^2-8xy-4y^2\)
\(=5x^2-26xy+5y^2==\left(5x-y\right)\left(x-5y\right)\)
a)\(\left(x^2+2x\right)^2+2\left(x^2+2x\right)+1\)
\(=\left(x^2+2x+1\right)^2=\left(x+1\right)^4\)
b) \(16x^2-9\left(x+y\right)^2\)
\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)
\(=\left(4x+3x+3y\right)\left(4x-3x-3y\right)\)
\(=\left(7x+3y\right)\left(x-3y\right)\)
\(\left(3x+1\right)^2-4\left(x-2\right)^2=9x^2+6x+1-4\left(x^2-4x+4\right)=9x^2+6x+1-4x^2+16x-16=5x^2+22x-15=\)
\(\left(5x-3\right)\left(x+5\right)\)
\(9\left(2x+3\right)^2-4\left(x+1\right)^2=9\left(4x^2+12x+9\right)-4\left(x^2+2x+1\right)=36x^2+108x+81-4x^2-8x-4=32x^2+100x+77\)
\(\left(8x+11\right)\left(4x+7\right)\)
\(x^4+2x^2y+y^2-9\)
\(=\left(x^2+y\right)^2-3^2\)
\(=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
\(x^4+2x^2y+y^2-9\)
\(=\left(x^2\right)^2+2.x^2.y+y^2-3^2\)
\(=\left(x^2+y\right)^2-3^2\)
\(=\left(x^2+y-3\right)\left(x^2+y+3\right)\)