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mk viết đáp án, ko biết biến đổi ib mk
a) \(x^3+3x^2y-9xy^2+5y^3=\left(x+5y\right)\left(x-y\right)^2\)
b) \(x^4+x^3+6x^2+5x+5=\left(x^2+5\right)\left(x^2+x+1\right)\)
c) \(x^4-2x^3-12x^2+12x+36=\left(x^2-6\right)\left(x^2-2x-6\right)\)
d) \(x^8y^8+x^4y^4+1=\left(x^2y^2-xy+1\right)\left(x^2y^2+xy+1\right)\left(x^4y^4-x^2y^2+1\right)\)
a )\(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-2y-2\right)\left(x+2y\right)\)
b )\(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+2x+2\right)\left(x^2-2\right)\)
c ) \(x^2\left(1-x^2\right)-4-4x^2=x^2-x^4-4-4x^2\)
\(=x^2-\left(x^2+2\right)^2=\left(x-x^2-2\right)\left(x^2+x+2\right)\)
\(x^3+4x^2+4x+3\)
\(=x^3+3x^2+x^2+3x+x+3\)
\(=x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+x+1\right)\)
\(x^2-y^2+4y-4\)
\(=x^2-\left(y^2-4y+4\right)\)
\(=x^2-\left(y-2\right)^2\)
\(=\left(x-y+2\right)\left(x+y-2\right)\)
\(x^4+x^3y-xy^3-y^4\)
\(=x^3\left(x+y\right)-y^3\left(x+y\right)\)
\(=\left(x+y\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
Chúc bạn học tốt.
a, \(x^4+6x^3+7x^2-6x+1\)
\(=x^4-2x^2+1+6x^3+9x^2+6x\)
\(=\left(x^2-1\right)^2+6x\left(x^2-1\right)+9x^2\)
\(=\left(x^2-1+3x\right)^2\)
b, \(x^4-7x^3+14x^2-7x+1\)
\(=x^4+2x^2+1+7x^3+12x^2-7x\)
\(=\left(x^2+1\right)^2-7x\left(x^2+1\right)+12^2\)
\(=\left(x^2-1+3x\right)^2\)
c, \(12x^2-11x-36\)
\(=12x^2-27x+16x-36\)
\(=3x\left(4x-9\right)+4\left(4x-9\right)\)
\(=\left(4x-9\right)\left(3x+4\right)\)
1, \(=\left(2y\right)^2-\left(x^2-2x+1\right)=\left(2y\right)^2-\left(x-1\right)^2=\left(2y-x+1\right)\left(2y+x-1\right)\)
2, \(=2\left(x^2-y^2\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1\right)+8\left(x+1\right)=2\left(x+1\right)\left(x-1+4\right)=2\left(x+1\right)\left(x+3\right)\)
3, \(=\left(x^2+6x+9\right)-\left(2y\right)^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
4, \(=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
\(4y^2-x^2+2x-1\)
\(=4y^2-\left(x^2-2x+1\right)\)
\(=\left(2y\right)^2-\left(x-1\right)^2\)
\(=\left(2y-x+1\right)\left(2y+x-1\right)\)
hk tốt
^^
a) nhận xét hệ số : 1 + 4 - 29 + 24 = 0
=> x3 + 4x2 - 29x + 24 = x2(x-1) + 5x(x-1) - 24(x-1)
= (x-1)(x2+5x-24) = (x-1)(x-3)(x+8)
b) ...
a) \(x^3+4x^2-29x+24\)=\(\left(x+8\right)\left(x^2-4x+3\right)\)=\(\left(x+8\right)\left(x^2-x-3x+3\right)\)=\(\left(x+8\right)\left(x-1\right)\left(x-3\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)=\(x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)=\(x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)=\(\left(x^2+3x-1\right)^2\)
\(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x+x+2\right)\left(x^2+x-x-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
\(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2-4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y-1\right)^2=\left(x-1+2y-1\right)\left(x-1-2y+1\right)\)
\(=\left(x-2y\right)\left(x+2y-2\right)\)
a) \(x^{12}-3x^6+1\)
\(=\left(x^6\right)^2-2\cdot x^6\cdot1+1^2-x^6\)
\(=\left(x^6-1\right)^2-\left(x^3\right)^2\)
\(=\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)\)
b) \(x^4+6x^3+7x^2-6x+1\)
\(=x^4+\left(6x^3-2x^2\right)+\left(9x^2-6x+1\right)\)
\(=\left(x^2\right)^2+2x^2\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x-1\right)^2\)
a) \(x^3+6x^2=x^2\left(x+6\right)\)
b) \(x^4+4y^4=1\left(x^4+4y^4\right)\)
a) x^3 + 6x^2 = x^2 ( x + 6)