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\(4x^2+6xy+y^2\)
\(=\left(9x^2+6xy+y^2\right)-8x^2\)
\(=\left(3x+y\right)^2-8x^2\)
\(=\left(3x+y-\sqrt{8}x\right)\left(3x+y+\sqrt{8}x\right)\)
\(4x^2+6xy+y^2\)
\(=\left(9x^2+6xy+y^2\right)-5x^2\)
\(=\left(3x+y\right)^2-5x^2\)
\(=\left(3x+y+\sqrt{5}x\right)\left(3x+y-\sqrt{5}x\right)\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
x4y4+64=x4y4+16x2y2+64-16x2y2
=(x2y2+8)2-16x2y2
=(x2y2-4xy+8)(x2y2+4xy+8)
\(x^4.y^4+4\)
\(=\left(x^4y^4-2x^3y^3+2x^2y^2\right)+\left(2x^3y^3-4x^2y^2+4xy\right)+\left(2x^2y^2-4xy+4\right)\)
\(=x^2y^2\left(x^2y^2-2xy+2\right)+2xy\left(x^2y^2-2xy+2\right)+2\left(x^2y^2-2xy+2\right)\)
= (x2y2 + 2xy + 2)(x2y2 - 2xy + 2)
Có: \(\left(x+y\right)^4+x^4+y^4\)
\(=\left(x+y\right)^4+\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left[\left(x+y\right)^4-x^2y^2\right]+\left[\left(x^2+y^2\right)^2-x^2y^2\right]\)
\(=\left[\left(x^2+y^2+2xy\right)^2-\left(xy\right)^2\right]+\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^2+y^2+xy\right)\left(x^2+y^2+3xy\right)+\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\)
\(=2\left(x^2+y^2+xy\right)\left(x^2+y^2+xy\right)=2\left(x^2+y^2+xy\right)^2\)
x4+4
=x4+4x2+4-4x2
=(x4+4x2+4)-(2x)2
=(x2+2)2-(2x)2
= (x2+2x+2).(x2-2x+2)
Ta có: \(x^4+6^4=x^4+72x^2+6^4-72x^2\)
\(=\left(x^2+36\right)-72x^2\)
\(=\left(x^2+36-\sqrt{72}x\right)\left(x^2+36+\sqrt{72}x\right)\)
Ta có : \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
\(x^4+4=\left(x^4+4x^2+4\right)-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)