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\(x^4-5x^2+4=x^4-x^2-4x^2+4=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-4\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
Ta có : x4 - 5x2 + 4
= x4 - x2 - 4x2 + 4
= x2(x2 - 1) + (4x2 - 4)
= x2(x2 - 1) + 4(x2 - 1)
= (x2 - 1)(x2 + 4)
#)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:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(a,x^2+9x+20=x^2+4x+5x+20.\)
\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
\(b,x^4-5x^2+4=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(c,x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2-2\right)-\left(2x\right)^2=\left(x^2-2x-2\right)\left(x^2+2x-2\right)\)
\(d,x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1\)
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)\left(x^2+3x\right)+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
\(x^4-5x^2+4=x^4-x^2-4x^2+4\)
\(=\left(x^2-1\right)\left(x^2+1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-4\right)=\left(x^2-1\right)\left(x^2-3\right)\)
a) \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)
b) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)
\(=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=\left(x-2\right)\left(8x-1\right)\)
\(3x\left(x-2\right)-x+2+5x\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)+5x\left(x-2\right)=\left(x-2\right)\left(3x=1+5x\right)=\left(x-2\right)\left(8x-1\right)\)
Ta có : x2 + 5x + 4
= x2 + x + 4x + 4
= x(x + 1) + 4(x + 1)
= (x + 1)(x + 4)