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a)=3xy2
b) x5+4x3-6x2 : 4x2
x5 : \(\overline{\frac{1}{4}x^3+x-1}\)
4x3-6x2 :
4x3 :
-6x2 :
-6x2 :
0
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Ta có:
\(2x^3-5x^2+6x-15\)
\(=\left(2x^3-5x^2\right)+\left(6x-15\right)\)
\(=x^2\left(2x-5\right)+3\left(2x-5\right)\)
\(=\left(x^2+3\right)\left(2x-5\right)\)
\(\Rightarrow\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)=x^2+3\)
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a) 3x^2-6x : -x+2
3x^2-6x : -3x
0
b) x^3 +2x^2 -2x -1 : x^2+3x+1
x^3 +3x^2 +x : x-1
-x^2 -3x -1 :
-x^2 -3x -1
0
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câu 1:
x3-1+3x2-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)
Câu 2 :
a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)
\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)
\(=\left[x^2\left(x^2-1\right)-2x\left(x^2-1\right)+\left(x^2-1\right)\right]:\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-2x+1\right):\left(x^2-1\right)\)
\(=x^2-2x+1\)
b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)
\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)
Câu 3 :
Sửa đề :
\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
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a ) \(\left(3x^2-4x+5\right)\left(2x^2-4\right)-2x\left(3x^3-4x^2+8\right)\)
\(=\left(3x^2-4x+5\right).2x^2-4\left(3x^2-4x+5\right)-6x^4+8x^3-16x\)
\(=6x^4-8x^3+10x^2-12x^2+16x-20-6x^4+8x^3-16x\)
\(=\left(6x^4-6x^4\right)+\left(8x^3-8x^3\right)-\left(12x^2-10x^2\right)+\left(16x-16x\right)-20\)
\(=-2x^2-20\)
b ) \(\left(1-3x+x^2\right)\left(2-4x\right)+2x\left(2x^2+5\right)\)
\(=2\left(1-3x+x^2\right)-4x\left(1-3x+x^2\right)+4x^3+10x\)
\(=2-6x+2x^2-4x+12x^2-4x^3+4x^3+10x\)
\(=\left(4x^3-4x^3\right)+\left(12x^2+2x^2\right)+\left(10x-6x-4x\right)+2\)
\(=14x^2+2\)
\(\left(-2x^5+6x^3-4x^2\right):2x^2\)
\(=-2x^5:2x^2+6x^3:2x^2-4x^2:2x^2\)
\(-x^3+3x-2\)