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\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
x4-2x3+2x-1
=x4-3x3+3x2-x+x3-3x2+3x-1
=x(x3-3x2+3x-1)+1(x3-3x2+3x-1)
=(x3-3x2+3x-1)(x+1)
=(x-1)3(x+1)
x4 + 2x3 + 2x2 + 2x + 1
= x4 - 2x2 =
= x2 x x2 - x2 - x2 + 1 = x2 (1- x2 ) + ( 1 - x2 )
= ( 1 - x2 ) x ( 1 - x2 )
= ( 1 - x2 ) 2
- SKT_Twisted Fate Âm Phủ
- Sai rồi :
- \(x^4-2x^2=?\)
=x^3 -8y^3 -2(x-2y)
=(x-2y)(x^2 +2xy +4y^2)- 2(x-2y)
=(x-2y)(x^2+2x +4y^2-2)
k day nhe
\(x^4+2x^3-4x-4\)
\(=x^4+2x^3-4x-4+2x^2-2x^2\)
\(=\left(x^4-2x^2\right)+\left(2x^3-4x\right)+\left(2x^2-4\right)\)
\(=x^2\left(x^2-2\right)+2x\left(x^2-2\right)+2\left(x^2-2\right)\)
\(=\left(x^2+2x+2\right)\left(x^2-2\right)\)
\(=\left(x^2+2x+2\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)
\(x^4-2x^3+2x-1\)
\(=\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)\)
\(=x\left(x-1\right)^3+\left(x-1\right)^3\)
\(=\left(x+1\right)\left(x-1\right)^3\)
x^4 - 2x^3 +2x -1
=x^4 - x^3 - x^3 + x^2 -x^2 +x +x -1
=x^3(x-1) - x^2(x - 1) -x(x - 1) + (x - 1)
=(X - 1)(X^3 - X^2 -X +1)
=(X-1){ X^2(x - 1) - (x-1) }
=(x-1){ (x-1)(X^2 - 1)}
=(x - 1)(x - 1)(x - 1)( x + 1)= (X - 1)^3(X - 1)
\(a,\left(2x+3\right)\left(2x-3\right)-\left(2x+1\right)^2\)
\(=4x^2-9-4x^2-4x-1\)
\(=-4x-10\)
\(=-2\left(2x+5\right)\)
b,Tương tự
2x.3 hay 2x mũ 3 z bn
2x3 bạn ạ, sorry