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\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b)Ta có: x2y+xy2+x+y=2010
<=>xy.x+xy.y+x+y=2010
<=>11x+11y+x+y=2010
<=>12(x+y)=2010
<=>x+y=167,5
=>(x+y)2=28056,25
<=>x2+y2+2xy=28056,25
<=>x2+y2=28034,25
\(\text{a) }x^4+64\)
\(=x^4+16x^2+64-16x^2\)
\(=\left(x^4+16x^2+64\right)-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2+4x+8\right)\left(x^2-4x+8\right)\)
\(\text{b) }4x^4+81y^4\)
\(=4x^4+36x^2y^2+81y^4-36x^2y^2\)
\(=\left(4y^4+36x^2y^2+81y^4\right)-36x^2y^2\)
\(=\left(2x^2+9y^2\right)^2-\left(6xy\right)^2\)
\(=\left(2x^2+9y^2+6xy\right)\left(2x^2+9y^2-6xy\right)\)
a. x4 + 64
= (x2)2 + 2x28 + 82 - 2x28
= (x2 + 8)2 - (4x)2
= (x2 + 8 + 4x)(x2 + 8 - 4x)
b. 4x4 + 81y4
= (2x2)2 + (9y2)2
Làm tới đây bí rồi bạn! Mà hình như làm gì có công thức a2 + b2
\(a,\)\(x^3-13x-12\)
\(=x^3-x-12x-12\)
\(=x\left(x^2-1\right)-12\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
a) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen
c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha
\(64x^4+y^4\)
\(=\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
\(x^5+x-1\)
\(=x^5+x^2-\left(x^2-x+1\right)\)
\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
\(x^3\left(x^2-7\right)^2-36x\)
\(=x.\left[x^2.\left(x^2-7\right)^2-36\right]\)
\(=x.\left[\left(x^3-7x\right)^2-6^2\right]\)
\(=x.\left(x^3-7x-6\right).\left(x^3-7x+6\right)\)
\(=x.\left(x+1\right)\left(x^2-x-6\right).\left(x-1\right).\left(x^2+x-6\right)\)
\(=x.\left(x+1\right).\left(x+2\right).\left(x+3\right).\left(x-1\right).\left(x-2\right).\left(x-3\right)\)
Ta có : \(x^3\left(x^2-7\right)^2-36x\)
= \(x^3\left(x^4-14x^2+49\right)-36x\)
= \(x\left(x^6-14x^4+49x^2-36\right)\)
= \(x\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)\)---- chỗ này tắt
= (x-3)(x-2)(x-1)x(x+1)(x+2)(x+3)