Ta có:

\(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\\ =\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{x^3-y^3}\\ =\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ =\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)

    \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\) \(=\dfrac{x^2+xy+y^2}{x^3-y^3}-\dfrac{3xy}{x^3-y^3}+\dfrac{\left(x-y\right)^2}{x^3-y^3}\)

\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{x^3-y^3}\)

\(=\dfrac{2x^2+2y^2-4xy}{x^3-y^3}\)

\(=\dfrac{2x^2-2xy-2xy+2y^2}{x^3-y^3}\)

\(=\dfrac{2x\left(x-y\right)-2y\left(x-y\right)}{x^3-y^3}\)

\(=\dfrac{\left(2x-2y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(=\dfrac{2x-2y}{x^2+xy+y^2}\)

Đặt 1/x=a; 1/y=b

Hệ phương trình trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{2}{3}\\\dfrac{1}{4}a+\dfrac{1}{3}b=\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=2\\15a+20b=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}15b+15b=30\\15b+20b=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-5b=18\\a+b=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{18}{5}\\a=\dfrac{64}{15}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{18}\\y=\dfrac{15}{64}\end{matrix}\right.\)

a.\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{8}{x}+\dfrac{5}{y}=1\end{matrix}\right.\)

\(ĐK:x;y\ne0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\) 

hpt trở thành:

\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\8a+5b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{18}\\b=\dfrac{1}{9}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{18}\\\dfrac{1}{y}=\dfrac{1}{9}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\) ( tm )

Vậy nghiệm hpt: \(\left\{{}\begin{matrix}x=18\\y=9\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\\2x+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1+y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

=>9x+4y=360 và 36/x-36/y=1/2

=>4y=360-9x và 36/x-36/y=1/2

=>y=90-2,25x và \(\dfrac{36}{x}-\dfrac{36}{90-2,25x}=\dfrac{1}{2}\)

=>\(\dfrac{3240-81x-36x}{x\left(90-2,25x\right)}=\dfrac{1}{2}\)

=>90x-2,25x^2=2(3240-117x)

=>-2,25x^2+90x-6840+234x=0

=>x=118,3 hoặc x=25,7

=>y=-176,175 hoặc y=32,175

còn cách làm khác không ạ?

a) \(ĐKXĐ:2x^2+6x+1\ge0\)

Với \(x\ge2\) pt cho trở thành :

\(2x^2+6x+1=x^2+4x+4\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow x=3\) ( do \(x\ge2\) )

Vậy pt có tập nghiệm \(S=\left\{3\right\}\)

\(a.\sqrt{2x^2+6x+1}=x+2\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\2x^2+6x+1=x^2+4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x^2+2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow x=1\\ \Rightarrow S=\left\{1\right\}\)

\(b.\) ĐKXĐ: \(y\ne0\)\(\left(I\right)\Rightarrow x+\dfrac{1}{y}=\dfrac{x}{y}+\dfrac{1}{y}\Leftrightarrow x=\dfrac{x}{y}\Leftrightarrow x\left(1-\dfrac{1}{y}\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\y=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{1}{2}\\x=1\end{matrix}\right.\left(TM\right)\Rightarrow S=\left\{\left(0;\dfrac{1}{2}\right);\left(1;1\right)\right\}\)

a: =>2/x+2/y=2 và 4/x-2/y=1

=>6/x=3 và 1/x+1/y=1

=>x=2 và 1/y=1-1/2=1/2

=>x=2; y=2

b: Đặt 1/x=a; 1/y=b

=>1/3a+1/3b=1/4 và 5/6a+b=2/3

=>a=1/2; b=1/4

=>x=2; y=4