a) \(x^4+2x^2-3=x^4-x^2+3x^2-3=\left(x^4-x^2\right)+\left(3x^2-3\right)=x^2\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x^2+3\right)=\left(x-1\right)\left(x+1\right)\left(x^2+3\right)\)
b) Đặt \(x^2+2x=a\)
\(\Rightarrow B=a\left(a+4\right)+3\) ( Đặt biểu thức trên là B)
= \(a^2+4a+3\)
=\(a^2+a+3a+3\)
=\(\left(a^2+a\right)+\left(3a+3\right)\)
=\(a\left(a+1\right)+3\left(a+1\right)\)
=\(\left(a+1\right)\left(a+3\right)\)
Thay \(a=x^2+2x\)
\(\Rightarrow\left(x^2+2x+1\right)\left(x^2+2x+3\right)\)
a) x4 + 2x2 - 3
= x4 + 2x2 + 1 - 4
= (x2 + 1)2 - 22
= (x2 + 1 - 2)(x2 + 1 + 2)
= (x2 - 1)(x2 + 3)
= (x - 1)(x + 1)(x2 + 3)
c) 2xy2 + 4xy + 2x - 2xz2 + 4xzt - 2xt2
= 2x(y2 + 2y + 1 - z2 + 2zt - t2)
= \(2x\left [ (y + 1)^{2} - (z -t)^{2} \right ]\)
= 2x(y - 1 - z + t)(y + 1 + z - t)
= 2x(y - z + t - 1)(y + z - t + 1)
câu b mk đang suy nghĩ
c) \(2xy^2+4xy+2x-2xz^2+4xzt-2xt^2=2x\left(y^2+2y+1-z^2+2zt-t^2\right)=2x\left[\left(y^2+2y+1\right)-\left(z^2-2zt+t^2\right)\right]=2x\left[\left(y+1\right)^2-\left(z-t\right)^2\right]=2x\left(y+1-z+t\right)\left(y+1-z-t\right)\)
a, \(x^4+2x^2-3\)\(=\left[\left(x^2\right)^2+2.x^2.1+1\right]-\left(1-3\right)\)
\(=\left(x^2+1\right)^2-\left(\sqrt{2}\right)^2\)
\(=\left(x^2+1-\sqrt{1}\right)\left(x^2+1+\sqrt{2}\right)\)
b,\(\left(x^2+2x\right)\left(x^2+2x+4\right)+3=x^4+2x^3+4x^2+2x^3+4x^2+8x\)
\(=\left(x^4+4x^3\right)+\left(8x^2+8x\right)=x\left(x^3+4x^2\right)+x\left(8x+8\right)\)
\(=x\left(x^3+4x^2+8x+8\right)\)
c,\(2xy^2+4xy+2x-2xz^2+4xzt-2xt^2\)
\(=\left(2xy^2+4xy\right)+\left(2x-2xz^2\right)+\left(4xzt-2xt^2\right)\)
\(=2x\left(y^2+2y\right)+2x\left(1-z^2\right)+2x\left(2zt-t^2\right)\)
\(=2x\left(y^2+2y+1-z^2+2zt-t^2\right)\)
\(=2x\left[\left(y^2+2y+1\right)-\left(z^2-2zt+t^2\right)\right]\)
\(=2x\left[\left(y+1\right)^2-\left(z-t\right)^2+\left(2y-t^2\right)\right]\)
\(=2x\left[\left(y+1-z-t\right)\left(y+1+z-t\right)\right]\)
