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\(=\dfrac{x^4-x^3+x^2-2x+1}{x^2+x+1}\)
\(=\dfrac{x^4+x^3+x^2-2x^3-2x^2-2x+2x^2+1}{x^2+x+1}\)
\(=x^2-2x+\dfrac{2x^2+1}{x^2+x+1}\)
\(1.3x^2-x-10=3x^2+5x-6x-10=x\left(3x+5\right)-2\left(3x+5\right)=\left(3x+5\right)\left(x-2\right)\) \(2.2x^2+3x-9=2x^2-3x+6x-9=x\left(2x-3\right)+3\left(2x-3\right)=\left(2x-3\right)\left(x+3\right)\) \(3.2x^4-5x-7=2x^4+2x-7x-7=2x\left(x^3+1\right)-7\left(x+1\right)=2x\left(x+1\right)\left(x^2-x+1\right)-7\left(x+1\right)=\left(x+1\right)\left(2x^3-2x^2+2x-7\right)\)
a)
\(3x^2-x-10\)
\(\Leftrightarrow3x^2-6x+5-10\)
\(\Leftrightarrow3x\left(x-2\right)+5\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)\)
b) Đề sai nha bạn, mình mạo phép sửa lại tí nghen!
\(2x^2+3x-9\)
<=>\(2x^2-3x+6x-9\)
\(\Leftrightarrow2x\left(x+3\right)-3\left(x+3\right)\)
\(\Leftrightarrow\left(x+3\right)\left(2x-3\right)\)
c)
\(2x^4-5x-7\)
\(\Leftrightarrow2x^4-2x^3-2x^2-7x+2x^3+2x^2+2x-7\)
\(\Leftrightarrow2x^4-5x-7x-7\)
\(\Leftrightarrow\left(x+1\right)\left(2x^3-2x^2+2x-7\right)\)
Chúc bạn học tốt! ^^
\(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)
\(=-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+\left(2x^4-3x^2-2\right)-4x^2\left(9x^2-24x+16\right)\)
\(=-6x^4-60x^3-150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)
\(=-40x^4+36x^3-218x^2+6x-11\)
(đã thử lại)
\(x^3-x^2-4x^2+8x^4\)
\(=x^3-5x^2+8x^4=x^2\left(x-5+8x^2\right)\)
\(-x^2-2x+15=-\left(x^2+2x+1-16\right)\)
\(=-\left[\left(x-1\right)^2-16\right]=-\left(x-5\right)\left(x+3\right)\)
\(3x^2+4x+1=4x^2+4x+1-x^2\)
\(=\left(2x+1\right)^2-x^2=\left(x+1\right)\left(3x+1\right)\)
1: \(=3\left[16x^2y^2-\left(x^2-2xy+y^2\right)\right]\)
\(=3\left[\left(4xy\right)^2-\left(x-y\right)^2\right]\)
\(=3\left(4xy-x+y\right)\left(4xy+x-y\right)\)
3: \(=4\left(x^4+x^2y^2-2y^4\right)\)
\(=4\left(x^4+2x^2y^2-x^2y^2-2y^4\right)\)
\(=4\left[x^2\left(x^2+2y^2\right)-y^2\left(x^2+2y^2\right)\right]\)
\(=4\left(x^2+2y^2\right)\left(x-y\right)\left(x+y\right)\)
4: \(=-\left(x^2+2x-15\right)=-\left(x+5\right)\left(x-3\right)\)
5: =(x-1)(3x-1)
\(=-6x^2\left(x^2+10x+25\right)-x^2+6x-9+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(=-6x^4-60x^3-150x^2-x^2+6x-9+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(=-4x^4-60x^3-154x^2+6x-11-36x^4+96x^3-64x^2\)
\(=-40x^4+36x^3-218x^2+6x-11\)
Ta có: 2x4 - 3x2 - 5 = 2x4 + 2x2 - 5x2 - 5 = 2x2(x2 + 1) - 5(x2 + 1)
= (x2 + 1)(2x2 - 5)
\(\Rightarrow\) (x2 + 1)(2x2 - 5) \(\div\) (x2 + 1) = 2x2 - 5
Vậy 2x4 - 3x2 - 5 \(\div\) (x2 + 1) = 2x2 - 5