b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

\(a,=\dfrac{\left(x-2\right)^2-\left(x+2\right)^2}{\left(x-2\right)^2\left(x+2\right)^2}:\dfrac{x-2+x+2}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{-8x}{\left(x-2\right)^2\left(x+2\right)^2}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{2x}=\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

\(b,=\dfrac{5x^2+26xy+5y^2+5x^2-26xy+5y^2}{x\left(x-5y\right)\left(x+5y\right)}\cdot\dfrac{\left(x-5y\right)\left(x+5y\right)}{x^2+y^2}\\ =\dfrac{10\left(x^2+y^2\right)}{x\left(x^2+y^2\right)}=\dfrac{10}{x}\)

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

a: \(5x^2y^4:10x^2y=\dfrac{1}{2}y^3\)

c: \(\left(-xy\right)^{10}:\left(-xy\right)^5=-x^5y^5\)

a: \(\left\{{}\begin{matrix}\dfrac{-5x+2y}{3}+5=\dfrac{y+27}{4}-2x\\\dfrac{x+1}{3}+y=\dfrac{6y-5x}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4\left(-5x+2y\right)+60=3\left(y+27\right)-24x\\7\left(x+1\right)+21y=3\left(6y-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y+60=3y+81-24x\\7x+7+21y=18y-15x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-20x+8y-3y+24x=21\\7x+21y-18y+15x=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x+5y=21\\22x+3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+15y=63\\110x+15y=-35\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-98x=98\\4x+5y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\5y=21-4x=21+4=25\end{matrix}\right.\)

=>x=-1 và y=5

b: \(\left\{{}\begin{matrix}\dfrac{1}{2}\left(x+2\right)\left(y+3\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(x-2\right)\left(y-2\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{2}\left(xy+3x+2y+6\right)-\dfrac{1}{2}xy=50\\\dfrac{1}{2}xy-\dfrac{1}{2}\left(xy-2x-2y+4\right)=32\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}xy+3x+2y+6-xy=100\\xy-\left(xy-2x-2y+4\right)=64\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x+2y=94\\2x+2y=60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=34\\2x+2y=60\end{matrix}\right.\)

=>x=34 và y=-4

c: \(\left\{{}\begin{matrix}\left(x+20\right)\left(y-1\right)=xy\\\left(x-10\right)\left(y+1\right)=xy\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy-x+20y-20=xy\\xy+x-10y-10=xy\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-x+20y=20\\x-10y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10y=30\\x-10y=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x=10y+10=30+10=40\end{matrix}\right.\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-2y\\x< >-\dfrac{y}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2}{x+2y}+\dfrac{1}{2x+y}=3\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{4}{x+2y}+\dfrac{2}{2x+y}=6\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{2x+y}=5\\\dfrac{4}{x+2y}-\dfrac{3}{2x+y}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+y=1\\\dfrac{4}{x+2y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=1\\x+2y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+y=1\\2x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=1\\x+2y=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=1-2y=1-\dfrac{2}{3}=\dfrac{1}{3}\end{matrix}\right.\)(nhận)

e: ĐKXĐ: x<>-1 và y<>-4

\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=\dfrac{27}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{9}\\y=-\dfrac{87}{19}\end{matrix}\right.\left(nhận\right)\)

h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)

Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)

\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)

Thay a,b:

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)

a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)

Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)

=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}x-y=2\\y-3z=2\\-3x-2y+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3\left(y+2\right)-2\left(3z+2\right)+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-6-6z-4+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3\left(3z+2\right)-5z=8\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-9z-6-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-14z=14\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\left(-1\right)+2=1\\y=3\left(-1\right)+2=-1\\z=-1\end{matrix}\right.\)

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