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a)\(\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
b)\(x\left(y+1\right)+\left(y+1\right)=\left(y+1\right)\left(x+1\right)\)
c)\(\left(x+y\right)^2-2\left(x+y\right)=\left(x+y\right)\left(x+y-2\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
1. 4-32x3
= 4.(1-8x3)
= 4.[13-(2x)3 ]
= 4.(1-2x).(1+2x+4x2)
2. b. \(\left(\frac{x}{xy-y^2}-\frac{2x-y}{xy-x^2}\right):\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left[\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(x-y\right)}\right]:\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left[\frac{x.x}{y\left(x-y\right).x}+\frac{\left(2x-y\right).y}{x\left(x-y\right).y}\right]:\left(\frac{x+y}{xy}\right)\)
\(=\left[\frac{x^2+2xy-y^2}{xy\left(x-y\right)}\right]:\left(\frac{x+y}{xy}\right)\)
\(=\left[\frac{-\left(x-y\right)^2}{xy\left(x-y\right)}\right].\frac{xy}{x+y}\)
\(=\frac{-\left(x-y\right)}{xy}.\frac{xy}{x+y}\)
\(=\frac{y-x}{x+y}\)
a, = (x^2+10x+25)-y62 = (x+5)^2-y^2 = (x+5-y).(x+5+y)
b, = xy.(x-y)
c, = (x-y).(x+y)+5.(x-y) = (x-y).(x+y+5)
k mk nha
a)\(x^2+2x+1=\left(x+1\right)^2\)
b)\(1-2y+y^2=\left(1-y\right)^2\)
hơ hơ ~ dễ thế này cơ mà!
a.x2+2x+1=x2+2x+12=(x+1)2=(x+1)*(x+1)
b.1-2y+y2=12-2y+y2=(y-1)2=(y-1)*(y-1)
a) x2 + xy - 5x - 5y
=x(x+y)-5(x+y)
=(x-5)(x+y)
b) x2 - y2 - 4x + 4
=(x2-4x+4)-y2
=(x-2)2-y2
=(x-2-y)(x-2+y)
a) x2 + xy - 5x - 5y
= x ( x + y ) - 5 ( x + y )
= ( x + y ) ( x - 5 )
b) x2 - y2 - 4x + 4
= ( x2 - 4x + 4 ) - y2
= ( x - 2 )2 - y2
= ( x - 2 + y ) ( x - 2 - y )
\(2x^2+xy-y^2=\left(x^2-xy\right)+\left(x^2-y^2\right)=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left[x+\left(x-y\right)\right]=\left(x-y\right)\left(x+x-y\right)=\left(x-y\right)\left(2x+y\right)\)