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a) \(\frac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}=\frac{9x^2+12x+4-x^2-4x-4}{x^3-x^2}=\frac{8x^2+8x}{x^3-x^2}=\frac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\frac{8\left(x+1\right)}{x-1}\)
b) \(\frac{x^4-1}{x^3+2x^2-x-2}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^3-x\right)+\left(2x^2-2\right)}=\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x+2\right)\left(x^2-1\right)}=\frac{x^2+1}{x+2}\)
c) \(\frac{x^2+7x+12}{x^2+5x+6}=\frac{\left(x^2+3x\right)+\left(4x+12\right)}{\left(x^2+3x\right)+\left(2x+6\right)}=\frac{\left(x+3\right)\left(x+4\right)}{\left(x++3\right)\left(x+2\right)}=\frac{x+4}{x+2}\)
d) \(\frac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}=\frac{\left(x^{10}-x^8\right)+\left(x^6-x^4\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\frac{x^8+x^4+1}{x^2+1}\)
a) 5x3 - 40 = 5( x3 - 8 ) = 5( x - 2 )( x2 + 2x + 4 )
b) x2z + 4xyz + 4y2z = z( x2 + 4xy + 4y2 ) = z( x + 2y )2
c) 4x2 - y2 - 6x + 3y = ( 4x2 - y2 ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )
d) x2 + 2x - 4y2 + 1 = ( x2 + 2x + 1 ) - 4y2 = ( x + 1 )2 - ( 2y )2 = ( x - 2y + 1 )( x + 2y + 1 )
e) 3x2 - 3y2 - 12x + 12y = 3( x2 - y2 - 4x + 4y ) = 3[ ( x2 - y2 ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )
f) x3 + 5x2 + 4x + 20 = x2( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x2 + 4 )
g) x3 - x2 - 25x + 25 = x2( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x2 - 25 ) = ( x - 1 )( x - 5 )( x + 5 )
a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)
c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)
\(=\left(x+1\right)^2-4y^2=\left(x+2y+1\right)\left(x-2y+1\right)\)
e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)
\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)
\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)
g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)
\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)
\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)
a) x(2x^2 -3) -x^2 (5x+1 ) + x^2
<=> 2x^3 -3x -5x^3 -x^2 +x^2
<=>3x^3 -3x
b) 3x(x-2) -5x(1-x)-8(x^2 -3)
=3x^2 -6x -5x +5x^2 -8x^2 +24
= -11x+24
a)\(6x^2-9xy\)
\(=3x\left(2x-3y\right)\)
b)\(x^2-y^2-3x+3y\)
\(=\left(x-y\right)\left(x+y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)
c)\(x^4-8x^2-9\)
\(=x^4+x^2-9x^2-9\)
\(=x^2\left(x^2+1\right)-9\left(x^2+1\right)\)
\(=\left(x^2-9\right)\left(x^2+1\right)\)
\(=\left(x+3\right)\left(x-3\right)\left(x^2+1\right)\)
d)\(x^4-4\left(x^2+5\right)-25\)
\(=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2+5\right)\left(x^2-5-4\right)\)
\(=\left(x^2+5\right)\left(x^2-9\right)\)
\(=\left(x^2+5\right)\left(x-3\right)\left(x+3\right)\)
\(A=\dfrac{2x^3-18x}{x^4-81}\\ A=\dfrac{2x\left(x^2-9\right)}{\left(x^2+9\right)\left(x^2-9\right)}\\ A=\dfrac{2x}{x^2+9}\)
\(B=\dfrac{x^2-x-20}{x^2-25}\\ B=\dfrac{x\left(x-5\right)+4\left(x-5\right)}{\left(x+5\right)\left(x-5\right)}\\ B=\dfrac{\left(x-5\right)\left(x+4\right)}{\left(x+5\right)\left(x-5\right)}\\ B=\dfrac{x+4}{x+5}\)
\(C=\dfrac{8xy-6x^2}{12y^2-9xy}\\ C=\dfrac{2x\left(4y-3x\right)}{3y\left(4y-3x\right)}\\ C=\dfrac{2x}{3y}\)
\(E=\dfrac{x^2+5x+6}{x^2-4}\\ E=\dfrac{x\left(x+2\right)+3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\\ E=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-2\right)\left(x+2\right)}\\ E=\dfrac{x+3}{x-2}\)