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a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
a) \(2x^2-5x-12\)
\(=2x^2-8x+3x-12\)
\(=2x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(2x+3\right)\)
b) \(x^3+5x^2+8x+4\)
\(=\left(x^3+3x^2+2x\right)+\left(2x^2+6x+4\right)\)
\(=x\left(x^2+3x+2\right)+2\left(x^2+3x+2\right)\)
\(=\left(x^2+3x+2\right)\left(x+2\right)\)
\(=\left(x^2+x+2x+2\right)\left(x+2\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left(x+2\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x+2\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
c) \(x^4+x^2+1\)
\(=\left(x^4-x^3+x^2\right)+\left(x^3-x^2+x\right)+\left(x^2-x+1\right)\)
\(=x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(a,x^3+9x^2+27x+27-27z^3\)
\(=\left(x+3\right)^3-\left(3z\right)^3\)
\(=\left(x+3-3z\right)\left(x^2+6x+9+3xz+9z+9z^2\right)\)
.........
\(b,\)
\(=\left(x+1\right)^2\left(x-3\right)+x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+3x+1\right)\)
\(c,\)
\(=x^2\left(x^2+10\right)-2x\left(x^2+10\right)\)
\(=x\left(x-2\right)\left(x+10\right)\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
\(C=x^4+100x^2+99x+100\)
\(=x^4-x+100x^2+100x+100\)
\(=x\left(x^3-1\right)+100\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+100\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+100\right)\)
Câu 2 em khai triển hằng đẳng thức và rút gọn là ra nhé
C=x4+100x2+99x+100
C= x4-x + 100x2+100x+100
C=x(x3-1)+100(x2+x+1)
C=x(x-1)(x2+x+1)+100(x2+x+1)
C=(x2+x+1)(x2-x+100)
1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2, \(x^2-10x+25=\left(x-5\right)^2\)
3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
4, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2) \(x^2-10x+25=\left(x-5\right)^2\)
3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)
4) \(x^2+4xy+4y^2=\left(x+2y\right)^2\)