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Bài làm:
đk: \(x\ne-3;x\ne1\)
Ta có: \(\frac{x^2+6x+9}{1-x}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)
\(=\frac{\left(x+3\right)^2}{-\left(x-1\right)}\cdot\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}\)
\(=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)
\(=-\frac{x^2-2x+1}{2x+6}\)
\(ĐKXĐ:\hept{\begin{cases}x\ne-3\\x\ne1\end{cases}}\)
\(\frac{x^2+6x+9}{1-x}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x+3\right)^2}{x-1}.\frac{\left(x-1\right)^3}{2\left(x+3\right)^3}=\frac{-\left(x-1\right)^2}{2\left(x+3\right)}\)
\(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
= \(\frac{3x\left(x-y\right)}{5.2.\left(x+y\right)\left(x-y\right)}-\frac{x\left(x+y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x^2-3xy-x^2-xy}{10\left(x^2-y^2\right)}\)
= \(\frac{3x\left(x-y\right)}{10\left(x^2-y^2\right)}\)
= \(\frac{3x}{10\left(x+y\right)}\)
1/(x^2+6x+9)-1/(x^2-6x+9)=(x-3)/(x-3)(x+3)-(x+3)/(x-3)(x+3)= -6/(x-3)(x+3)
1/(x+3)+1/(x-3)=
a) \(\left(3-2x\right)\left(x+1\right)+x\left(2x-1\right)=3x+3-2x^2-2x+2x^2-x=3\)
b) \(\frac{x^2+9}{x^2+3x}+\frac{6}{x+3}=\frac{x^2+9}{x\left(x+3\right)}+\frac{6x}{x\left(x+3\right)}=\frac{x^2+6x+9}{x\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x+3\right)}=\frac{x+3}{x}\)
c)\(\frac{2+x}{2-x}+\frac{4x^2}{4-x^2}+\frac{x-2}{2+x}=\frac{\left(x+2\right)^2}{\left(2-x\right)\left(2+x\right)}+\frac{4x^2}{\left(2-x\right)\left(2+x\right)}+\frac{-\left(x-2\right)^2}{\left(2+x\right)\left(2-x\right)}\)
\(=\frac{x^2+4x+4+4x^2-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}=\frac{4x^2+8x}{\left(x+2\right)\left(2-x\right)}=\frac{4x\left(x+2\right)}{\left(x+2\right)\left(2-x\right)}=\frac{4x}{2-x}\)
d) \(\left(x^3+4x^2+6x+4\right):\left(x+2\right)\)
\(=\left(x^3+2x^2+2x^2+4x+2x+4\right):\left(x+2\right)\)
\(=\left[x^2\left(x+2\right)+2x\left(x+2\right)+2\left(x+2\right)\right]:\left(x+2\right)\)
\(=\left(x^2+2x+2\right)\left(x+2\right):\left(x+2\right)=x^2+2x+2\)
\(\frac{6x^3\left(2y+1\right)}{5y}\cdot\frac{15}{2x^3\left(2y+1\right)}=\frac{9}{y}\)
\(\frac{3}{x^2-1}:\frac{6x}{2x^3\left(2y+1\right)}=\frac{3}{x^2-1}\cdot\frac{2x^3\left(2y+1\right)}{6x}=\frac{x^2\left(2y+1\right)}{x^2-1}\)
hok tốt.
\(\frac{6x^3\left(2y+1\right)}{5y}\cdot\frac{15}{2x^3\left(2y+1\right)}\)
\(=\frac{6x^3\left(2y+1\right)}{5y}\cdot\left[\frac{15}{2x^3\left(2y+1\right)}\right]\)
\(=\frac{180x^3y+90x^3}{20x^3y^2+10x^3y}\)
\(=\frac{180y+90}{20y^2+10y}\)
\(=\frac{18y+9}{2y^2+y}\)
\(=\frac{9\left(2y+1\right)}{y\left(2y+1\right)}\)
\(=\frac{9}{y}\)