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a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
a) x2(5x3 – x - \(\frac{1}{2}\)) = x2. 5x3 + x2 . (-x) + x2 . ( \(-\frac{1}{2}\) )
= 5x5 – x3 – \(\frac{1}{2}\)x2
b) (3xy – x2 + y) \(\frac{2}{3}\)x2y = \(\frac{2}{3}\)x2y . 3xy + \(\frac{2}{3}\)x2y . (- x2) + \(\frac{2}{3}\)x2y .
y = 2x3y2 – \(\frac{2}{3}\)x4y + \(\frac{2}{3}\)x2y2
c) (4x3– 5xy + 2x)( \(-\frac{1}{2}\)xy) = \(-\frac{1}{2}\)xy . 4x3 + ( \(-\frac{1}{2}\)xy) . (-5xy) + ( \(-\frac{1}{2}\)xy) . 2x
= -2x4y + \(\frac{5}{2}\)x2y2 – x2y.
a ) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
b ) \(\left(x^2-2xy+y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x-y\right)=\left(x-y\right)^3\)
c ) \(\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\left(x-3y\right)\)
\(=\left(x^2y^2-\dfrac{1}{3}xy+3y\right)x-3y\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\)
\(=x^3y^2-\dfrac{1}{3}x^2y+3xy-3x^2y^3+xy^2-9y^2\)
d ) \(\left(\dfrac{1}{5}x-1\right)\left(x^2-5x+2\right)\)
\(=\dfrac{1}{5}x\left(x^2-5x+2\right)-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-x^2+\dfrac{2}{5}x-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-2x^2+\dfrac{27}{5}x-2\)
a)\(\left(3x^2+2xy\right).\left(5xy^2-4x+\frac{1}{3}y^2\right)\)
\(=15x^3y^2-12x^3+x^2y^3+10x^2y^3-8x^2y+\frac{2}{3}xy^4\)
\(=15x^3y^2-12x^3+11x^2y^3-8x^2y+\frac{2}{3}xy^4\)
b)\(\left(x^3-x^2-7x+3\right):\left(x-3\right)\)
\(=\left(x^3-3x^2+2x^2-6x-x+3\right):\left(x-3\right)\)
\(=\left[x^2\left(x-3\right)+2x\left(x-3\right)-\left(x-3\right)\right]:\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+2x-1\right):\left(x-3\right)\)
\(=x^2+2x-1\)
Bài 2:
\(=\dfrac{x^2\left(x^2+4\right)-2x\left(x^2+4\right)}{x^2+4}=x^2-2x\)
Bài 1:
a: \(=\left(\dfrac{2}{3}:\dfrac{-1}{9}\right)\cdot x^4y^2z^6=-6x^4y^2z^6\)
b: \(=-12x^8-21x^5\)
c: =x^3+8
d: \(=125x^3-75x^2+15x-1\)
Câu 1:
Sửa đề: \(\left(2x+1\right)^3+\left(x-5\right)^3+\left(-3x+4\right)^3=0\)
\(\Leftrightarrow\left(2x+1\right)^3+\left(x-5\right)^3-\left(3x-4\right)^3=0\)
Đặt a=2x+1; b=x-5
Phương trình sẽ là \(a^3+b^3-\left(a+b\right)^3=0\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-5\right)\left(3x-4\right)=0\)
hay \(x\in\left\{-\dfrac{1}{2};5;\dfrac{4}{3}\right\}\)
\(a,\left(4x^3-3xy^2+2xy\right).\left(-\dfrac{1}{3}x^2y\right)\)
\(=\dfrac{-x}{3y}+\dfrac{y}{x}-\dfrac{2}{3x}\)
\(b,\left(5xy-x^2+y\right)\left(\dfrac{2}{5}xy^2\right)\)
\(=\dfrac{2}{y}-\dfrac{2x}{5y^2}+\dfrac{2}{5xy}\)
c,=\(\dfrac{-4}{3}x^5y+x^3y^3-\dfrac{2}{3}x^3y^2\) b,=\(2x^2y^3-\dfrac{2}{5}x^3y^2+\dfrac{2}{5}xy^3\)