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\(x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
\(x^4+2x^3+2x^2+2x+1=x^4+x^2+2x^3+x^2+2x+1\)
\(=x^2\left(x^2+1\right)+2x\left(x^2+1\right)+\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+2x+1\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
\(x^4-2x^3+2x-1=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+1-2x\right)=\left(x^2-1\right)\left(x-1\right)^2\)
\(x^3+2x^2+2x+1=\left(x^3+x^2\right)+\left(x^2+x\right)+\left(x+1\right)\)
\(=x^2.\left(x+1\right)+x.\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right).\left(x^2+x+1\right)\)
\(x^3-4x^2+12x-27\)
\(=\left(x^3-x^2\right)-\left(3x^2-3x\right)+\left(9x-27\right)\)
\(=x^2.\left(x-1\right)-3x.\left(x-1\right)+9.\left(x-3\right)\)
\(=\left(x-1\right).\left(x^2-3x\right)+9.\left(x-3\right)\)
\(=x.\left(x-1\right).\left(x-3\right)+9.\left(x-3\right)\)
\(=\left(x-3\right)\left[x.\left(x-1\right)+9\right]\)
Nghịch xíu :v
a, \(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+2\right)\)
b, \(x^2+4x+3\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Chúc bạn học tốt!!!
=x3(x+2)-13x2+12x-26x+24
=x3(x+2)-x(13x-12)-2(13x-12)
=x3(x+2)-(13x-12)(x+2)
=(x+2)(x3-x-12x+12)
(x+2)[(x2-1)-12(x-1)]
=(x+2)[x(x-1)(x+1)-12(x-1)]
=(x+2)(x-1)[x(x+1)-12]
=(x+2)(x-1)(x2+x-12)
=(x+2)(x-1)(x2-3x+4x-12)
=(x+2)(x-1)[x(x-3)+4(x+3)]
=(x+2)(x-1)(x-3)(x+4)
trong bài làm của mk có hàng k có dấu "=" chỗ đó có dâu"=" nha!
a) \(x^4-2x^3+2x-1\)
\(=x^4-x^3-x^3+2x-2+1\)
\(=\left(x^4-x^3\right)+\left(2x-2\right)-\left(x^3-1\right)\)
\(=x^3\left(x-1\right)+2\left(x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x^3+2-x^2-x-1\right)\)
\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)
\(=\left(x-1\right)\left[\left(x^3-x^2\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x^2-1\right)\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3\left(x+1\right)\)
b) \(x^4+2x^3+2x^2+2x+1\)
\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)\)
\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+2x\right)\)
\(=\left(x^2+1\right)\left(x+1\right)^2\)
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 ) \(x^2\left(1-x^2\right)-4-4x^2=x^2-x^4-4-4x^2\)
\(=x^2-\left(x^2+2\right)^2=\left(x-x^2-2\right)\left(x^2+x+2\right)\)
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right).\left[x^2+4x-3x-12\right]\)
\(=\left(x-2\right).\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
\(x^4+x^3+2x-4\)
\(=x^4-x^3+2x^3-2x^2+2x^2-2x+4x-4\)
\(=x^3\left(x-1\right)+2x^2\left(x-1\right)+2x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+2x+4\right)\)
\(=\left(x-1\right).\left[x^2\left(x+2\right)+2\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+2\right)\)
\(8x^4-2x^3-3x^2-2x-1\)
\(=8x^4-8x^3+6x^3-6x^2+3x^2-3x+x-1\)
\(=8x^3\left(x-1\right)+6x^2\left(x-1\right)+3x\left(x-1\right)+x-1\)
\(=\left(x-1\right)\left(8x^3+6x^2+3x+1\right)\)
\(=\left(x-1\right)\left[\left(8x^3+1\right)+\left(6x^2+3x\right)\right]\)
\(=\left(x-1\right)\left[\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\right]\)
\(=\left(x-1\right)\left(2x+1\right)\left(4x^2+x+1\right)\)
\(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
Chúc bạn học tốt.
a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)
=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
=\(\left(x-1\right)\left(x^3+x^2-x+1\right)\)
b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)
=\(a^2b^2-2ab+1+a^2+2ab+b^2\)
=\(a^2b^2+a^2+b^2+1\)
=\(a^2\left(b^2+1\right)+\left(b^2+1\right)\)
=\(\left(b^2+1\right)\left(a^2+1\right)\)
c,\(x^4+2x^3+2x^2+2x+1\)
=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)
=\(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)
=\(\left(x+1\right)\left(x^3+x^2+x+1\right)\)
=\(\left(x+1\right)^2\left(x^2+1\right)\)
1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
\(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.
Vậy \(x^4+x^3+2x^2+x+1=\left(x^2+1\right)\left(x^2+x+1\right)\).