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\(\left(x+1\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)\)
\(\Leftrightarrow2.2x=4x\)
p/s tham khảo nha
\(a^2-b^2-a+b\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)-\left(a-b\right)\)
\(\Leftrightarrow\left(a-b\right)\left(a+b-1\right)\)
p/s tham khảo
a) \(x+x^2-x^3-x^4=x\left(1+x-x^2-x^3\right)\)
b) \(\left(x+1\right)^2-x-1=\left(x+1\right)^2-\left(x+1\right)=\left(x+1\right)\left(x+1-1\right)=x\left(x+1\right)\)
c) \(x^2-2x+1-y^2+2y-1=\left(x-1\right)^2-\left(y-1\right)^2=\left(x-1+y-1\right)\left(x-1-y+1\right)\)
\(=\left(x+y-2\right)\left(x-y\right)\)
d) \(3xy-z-3x+yz=3x\left(y-1\right)-x\left(y-1\right)=2x\left(y-1\right)\)
e) \(x^4-1-3\left(x^2+1\right)=\left(x^2-1\right)\left(x^2+1\right)-3\left(x^2+1\right)=\left(x^2+1\right)\left(x^2-1-3\right)\)
\(=\left(x^2+1\right)\left(x^2-4\right)=\left(x^2+1\right)\left(x-2\right)\left(x+2\right)\)
a, \(x+x^2-x^3-x^4=-x\left(x+1\right)^2\left(x-1\right)\)
b, \(\left(x+1\right)^2-x-1=x^2+2x+1-x-1=x^2+x=x\left(x+1\right)\)
c, \(x^2-2x+1-y^2+2y-1=\left(x-1\right)^2-\left(y-1\right)^2\)để thế này đc thôi
d, \(3xy-z-3x+yz=2x\left(y-1\right)\)
e, \(x^4-1-3\left(x^2+1\right)=x^4-1-3x^2-4=\left(x^2+1\right)\left(x-2\right)\left(x+2\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
a, \(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
a )
b)
c) x^5 - x^4 - 1
= x^5 - x^3 - x² - x^4 + x² + x + x^3 - x - 1
= x²( x^3 - x - 1 ) - x( x^3 - x - 1 ) + ( x^3 - x - 1 )
= ( x² - x + 1)( x^3 - x - 1 )
d)
a) \(x^{12}-3x^6+1\)
\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)
\(=\left(x^6-1\right)^2-\left(x^3\right)^2\)
\(=\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)