a) x3y-2x2y2+5xy
b) x.(y-x)3-y.(x-y)2+xy.(x-y)
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1: \(A=2x^3y^4-5x\cdot x^2y^4+xy^2\cdot x^2y^2=-2x^3y^4=-2\cdot\left(-1\right)^3\cdot\dfrac{1}{16}=\dfrac{1}{8}\)
2: \(B=9x^4y^6\cdot\left(-4xy\right)+19x^3y^5\cdot\left(-2\right)x^2y^2\)
\(=-36x^5y^7-38x^5y^7\)
\(=-74x^5y^7=-74\cdot\left(-1\right)^5\cdot2^7=9472\)
3: \(f\left(-1\right)=3\cdot\left(-1\right)^4+7\cdot\left(-1\right)^3+4\cdot\left(-1\right)^2-2\cdot\left(-1\right)-2=0\)
\(f\left(1\right)=3+7+4-2-2=10\)
4A:
a: \(A=a\left(b+3\right)-b\left(b+3\right)\)
\(=\left(b+3\right)\left(a-b\right)\)
\(=2000\cdot6=12000\)
b: \(B=b^2-8b-c\left(8-b\right)\)
\(=b\left(b-8\right)+c\left(b-8\right)\)
\(=\left(b-8\right)\left(b+c\right)\)
\(=100\cdot100=10000\)
c: Ta có: x=16
nên x+1=17
Ta có: \(C=x^4-17x^3+17x^2-17x+20\)
\(=x^4-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+20\)
\(=x^4-x^4-x^3+x^3+x^2-x^2-x+20\)
=20-x
=4
a) \(\left(2x^3-x^2+5x\right):x\)
\(=\dfrac{2x^3-x^2+5x}{x}\)
\(=\dfrac{x\left(2x^2-x+5\right)}{x}\)
\(=2x^2-x+5\)
b) \(\left(3x^4-2x^3+x^2\right):\left(-2x\right)\)
\(=\dfrac{3x^4-2x^3+x^2}{-2x}\)
\(=\dfrac{2x\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)}{-2x}\)
\(=-\left(\dfrac{3}{2}x^3-x^2+\dfrac{1}{2}x\right)\)
\(=-\dfrac{3}{2}x^3+x^2-\dfrac{1}{2}x\)
c) \(\left(-2x^5+3x^2-4x^3\right):2x^2\)
\(=\dfrac{-2x^5+3x^2-4x^3}{2x^2}\)
\(=\dfrac{2x^2\left(-x^3+\dfrac{3}{2}-2x\right)}{2x^2}\)
\(=-x^3-2x+\dfrac{3}{2}\)
d) \(\left(x^3-2x^2y+3xy^2\right):\left(-\dfrac{1}{2}x\right)\)
\(=\dfrac{x^3-2x^2y+3xy^2}{-\dfrac{1}{2}x}\)
\(=\dfrac{\dfrac{1}{2}x\left(2x^2-4xy+6y^2\right)}{-\dfrac{1}{2}x}\)
\(=-\left(2x^2-4xy+6y^2\right)\)
\(=-2x^2+4xy-6y^2\)
e) \(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:5\left(x-y\right)^2\)
\(=\dfrac{3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2}{5\left(x-y\right)^2}\)
\(=\dfrac{5\left(x-y\right)^2\left[\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\right]}{5\left(x-y\right)^2}\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
f) \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2\)
\(=\dfrac{3x^5y^2+4x^3y^3-5x^2y^4}{2x^2y^2}\)
\(=\dfrac{2x^2y^2\left(\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\right)}{2x^2y^2}\)
\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)
phân tích đa thức thành nhân tử hả?
a) tự chép lại đề
=xy(x2-2xy+5)