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a,
$xy^2+x^2y+(-2xy^2)=xy^2-2xy^2+x^2y=-xy^2+x^2y$
b,
$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$
c,
$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$
d,
$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$
$=3x^2+x$
e,
$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$
f,
$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$
$=\frac{23}{4}xy^2$
\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
\(a.\dfrac{x+1}{2x+6}+2x=\dfrac{x+1+4x^2+12x}{2x+6}=\dfrac{4x^2+13x+1}{2x+6}\) ( x # -3)
\(b.\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\) ( x # - 3)
Các câu còn lại tương tự .
\(a,\dfrac{x+1}{2x+6}+2x\)
\(=\dfrac{x+1}{2x+6}+\dfrac{2x\left(2x+6\right)}{2x+6}\)
\(=\dfrac{x+1+4x^2+12x}{2x+6}\)
\(=\dfrac{4x^2+13x+1}{2x+6}\)
\(b,\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3x}{2x^2+6x}-\dfrac{x-6}{2x^2-6x}\)
\(=\dfrac{2x-6}{2x^2+6x}=\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}=\dfrac{x-3}{x^2+3x}\)
\(c,\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)
\(=\dfrac{x\left(x+2y\right)}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{x\left(x-2y\right)}{\left(x+2y\right)\left(x-2y\right)}-\dfrac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x}{x+2y}\)
\(d,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{4-9x^2}\)
\(=\dfrac{3x+2}{\left(3x+2\right)\left(3x-2\right)}-\dfrac{3x-2}{\left(3x+2\right)\left(3x-2\right)}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)
a: \(=2x^6-x^5-\dfrac{1}{3}x^4\)
b: \(=4xy^2-x^3+y^2-\dfrac{3}{4}x^2y\)
c: \(\left(3x^3-2xy^3+4y^2\right)\cdot\left(\dfrac{1}{6}x^2y^2\right)\)
\(=\dfrac{1}{2}x^5y^2-\dfrac{1}{3}x^3y^5+\dfrac{2}{3}x^2y^4\)