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4x2+4x-3
= 4x2+4x+1-4
=(2x+1)2-22
=(2x+1-2)(2x+1+2)
=(2x-1)(2x+3)
\(4x^2+4x-3\)
\(= 4x^2 + 4x + 1 -4\)
\(= ( 2x+1)^2 -2^2\)
\(= (2x+1-2)(2x+1+2)\)
\(= (2x-1)(2x+3)\)
\(x^3+4x^2+4x+1\)
\(=x^3+3x^2+x+x^2+3x+1\)
\(=x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)\)
\(=\left(x+1\right)\left(x^2+3x+1\right)\)
\(4x^2+4x-3=\left(2x\right)^2+2.\left(2x\right).1+1^2-4=\left(2x+1\right)^2-2^2=\left(2x+1-2\right)\left(2x+1+2\right)=\left(2x-1\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^3-x\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left[\left(2x\right)^2-1\right]\)
\(=x\left(x+1\right)\left(2x+1\right)\left(2x-1\right)\)
(Nhớ k cho mình với nhá!)
\(x^3-4x^2-xy^2+4x\)
\(=x\left(x^2-4x-y^2+4\right)\)
\(=x\left(\left(x-2\right)^2-y^2\right)\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
x3−4x2−xy2+4xx3−4x2−xy2+4x
=x(x2−4x−y2+4)=x(x2−4x−y2+4)
=x((x−2)2−y2)=x((x−2)2−y2)
=x(x−2−y)(x−2+y)
\(x^4+4x^3+2x^2-4x+1\)
\(=x^4+2x^3-x^2+2x^3+4x^2-2x-x^2-2x+1\)
\(=x^2\left(x^2+2x-1\right)+2x\left(x^2+2x-1\right)-\left(x^2+2x-1\right)\)
\(=\left(x^2+2x-1\right)^2\)
A,
x^2 - y^2 -2x -2y
= (x^2 - y^2) -(2x +2y)
= (x+y)(x-y) -2(x+y)
= (x+y)(x-y-2)
B,
5x^6 - 320
=5(x^6 - 64)
=5( (x^3)^2 - 8^2)
= 5( x^3 - 8)(x^3+8)
=5(x-2)(x^2 + 2x+4)(x+2)(x^2-2x-4)
\(4x^2+4x-3=\left(2x-1\right)\left(2x+3\right)\)
Ta có \(4x^2+4x-3=4x^2-2x+6x-3\)
\(=2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x+3\right)\)
Vậy \(4x^2+4x-3=\left(2x-1\right)\left(2x+3\right)\)