Phân tích đa thức thành nhân tử bằng cách áp dụng nhiều phương pháp
1) x2 - 2x - 4y2 - 4y
2) x4 + 2x3 - 4x - 4
3) x2 . (1 - x2) - 4 + 4x2
Giải từng bước một giúp e nhé. E cảm ơn nhiều ạ!
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1/ \(3x^2+6x+3-3y^2=3x^2+3x+3x+3-3y^2\)
\(=3\left(x^2+2x+1-y^2\right)\)
\(=3\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=3\left[\left(x+1\right)^2-y^2\right]\)
\(=3\left(x+1-y\right)\left(x+1+y\right)\)
2/ \(25-x^2-y^2+2xy=5^2-\left(x^2+y^2-2xy\right)\)
\(=5^2-\left(x-y\right)^2\)
\(=\left[5-\left(x-y\right)\right]\left(5+x+y\right)\)
\(=\left(5-x+y\right)\left(5+x+y\right)\)
3/ \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3-\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3-x+y\right)\)
a) \(x^2\left(x-3\right)+12-4x=x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x^2-2^2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x+2\right)\)
b)\(x^2-4+\left(x-2\right)^2=x^2-2^2+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2+x-2\right)\)
\(=\left(x-2\right)2x\)
c)\(x^3-4x^2-12x+27=x^3+3x^2-7x^2-21x+9x+27\)
\(=x^2\left(x+3\right)-7x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
a) => x2.(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
b) => (x+2)(x-2)+(x-2)2=(x-2)(x+2+x-2)=2x(x-2)
c) => x3+27-(4x2+12x)=(x+3)(x2-3x+3)-4x(x+3)=(x+3)(x2-3x+3-4x)=(x-3)(x2-7x+3)
a)\(x^4-2x^3+2x-1=x^4-x^3-x^3+x+x-1\)
\(=x^3\left(x-1\right)-x\left(x^2-1\right)+\left(x-1\right)\)
\(=x^3\left(x-1\right)-x\left(x-1\right)\left(x+1\right)+\left(x-1\right)\)
\(=x^3\left(x-1\right)-\left(x^2+x\right)\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left[x^3-\left(x^2+x\right)+1\right]\)
\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x-1\right)\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3\left(x+1\right)\)
b)\(x^4+2x^3+2x^2+2x+1=x^4+x^3+x^3+x^2+x^2+x+x+1\)
\(=x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x^2+x+1\right)\)
\(=\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=\left(x+1\right)^2\left(x^2+1\right)\)
1) \(x^2-2x-4y^2-4y\)
\(=\left[x^2-\left(2y\right)^2\right]-\left(2x+4y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
2) \(x^4+2x^3-4x-4\)
\(=\left(x^4-4\right)+\left(2x^3-4x\right)\)
\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2+2x\right)\)
3) \(x^2\left(1-x^2\right)-4x+4x^2\)
\(=x^2\left(1+x\right)\left(1-x\right)+4x\left(x-1\right)\)
\(=x^2\left(1+x\right)\left(1-x\right)-4x\left(1-x\right)\)
\(=\left(1-x\right)\left[x^2\left(1+x\right)-4x\right]\)