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a)x3+x2+4
=x3-x2+2x+2x2-2x+4
=x(x2-x+2)+2(x2-x+2)
=(x+2)(x2-x+2)
b)x3-2x-4
=x3+2x2+2x-2x2-4x-4
=x(x2+2x+2)-2(x2+2x+2)
=(x-2)(x2+2x+2)
Trả lời:
1) sửa đề: \(x^4+x^3-4x-4=x^3\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^3-4\right)\)
2) \(x^2-\left(a+b\right)x+ab=x^2-ax-bx+ab=\left(x^2-ax\right)-\left(bx-ab\right)\)
\(=x\left(x-a\right)-b\left(x-a\right)=\left(x-a\right)\left(a-b\right)\)
3) \(5xy^3-2xyz-15y^2+6z=\left(5xy^3-15y^2\right)-\left(2xyz-6z\right)\)
\(=5y^2\left(xy-3\right)-2z\left(xy-3\right)=\left(xy-3\right)\left(5y^2-2z\right)\)
\(x^4+16\)
\(=x^4+4x^2+16-4x^2\)
\(=\left(x^2+4\right)^2-4x^2\)
\(=\left(x^2-2x+4\right)\left(x^2+2x+4\right)\)
\(x^4+16\)
\(=x^4+4x^2+16-4x^2\)
\(=\left(x^2+4\right)^2-4x^2\)
\(=\left(x^2-2x+4\right)\left(x^2+2x+4\right)\)
\(=x^4+2x^2+1-\left(\sqrt{2}x\right)^2\)
\(=\left(x^2+1\right)^2-\left(\sqrt{2}x\right)^2\)
\(=\left(x^2+1-\sqrt{2}x\right)\left(x^2+1+\sqrt{2}x\right)\)
\(x^4+1\)
\(=x^4+2x^2+1-2x^2\)
\(=\left(x^2+1\right)^2-\left(x\sqrt{2}\right)^2\)
\(=\left(x^2-x\sqrt{2}+1\right)\left(x^2+x\sqrt{2}+1\right)\)
\(x^4+x^2+x\)
\(=x^2+x+x^4\)
\(=x^2+2x.\frac{1}{2}+\frac{1}{2}^2-\frac{1}{2}^2+x^4\)
\(=\left(x^2+2x.\frac{1}{2}+\frac{1}{2}^2\right)-\frac{1}{2}^2+x^4\)
\(=\left(x+\frac{1}{2}\right)^2-\frac{1}{2}^2+x^4\)
\(=\left(x+\frac{1}{2}\right)^2-\frac{1}{4}+x^4\)
\(=\left(x+\frac{1}{2}\right)^2-\sqrt{\frac{1}{4}}^2+x^4\)
\(=\left(x+\frac{1}{2}-\sqrt{\frac{1}{4}}\right).\left(x+\frac{1}{2}+\sqrt{\frac{1}{4}}\right)+x^4\)
Đến đây dễ rồi .Biến đổi ngoặc bên phải giống ngoặc trái rồi mở ngoặc đặt nhân tử chung là được .
\(x^3+3x^2-4\)
\(=\left(x^3+4x^2\right)-\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\)
Mình nhìn nhầm đề
\(x^3+3x^2-4\)
\(=\left(x^3+2x^2\right)+\left(x^2-4\right)\)
\(=x^2\left(x+2\right)+\left(x-2\right)\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+x-2\right)\)
\(=\left(x+2\right)\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)
\(=\left(x+2\right)\left(x+2\right)\left(x-1\right)\)
\(=\left(x+2\right)^2\left(x-1\right)\)
\(2x^4+128y^4\)
\(=2x^4+2.\left(8y^2\right)^2\)
\(=2\left[x^4+\left(8y^2\right)^2\right]\)
\(=2\left[x^4+2x^28y^2+\left(8y^2\right)^2-2x^28y^2\right]\)
\(=2\left[\left(x^2+8y^2\right)^2-\left(4xy\right)^2\right]\)
\(=2\left(x^2-4xy+8y^2\right)\left(x^2+4xy+8y^2\right)\)
2x4+128x^4
2x^4+2.(8y^2)^2
2.(x^4+(8y^2)^2)
2.((x^2)^2+2.x^2.8y^2+(8y^2)^2-2x^2.8y^2)
2.(x^2+8y^2)-(4.x.y)^2
2.((x^2+8y^2)-4xy).((x^2+8y^2)+4xy)
2.(x^2+8y^2-4xy).(x^2+8y^2+4xy)