a: x-2y=5 và 3x+y=8

=>3x-6y=15 và 3x+y=8

=>-7y=7 và x-2y=5

=>y=-1 và x=5+2y=5-2=3

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{6}{y-2}=9\\\dfrac{3}{x+1}-\dfrac{1}{y-2}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y-2}=7\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=3\end{matrix}\right.\)

=>y-2=1 và x+1=1

=>x=0 và y=3

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3.4}{x+y}+\dfrac{3.4}{x-y}=\dfrac{5}{2}\\\dfrac{4}{x+y}+\dfrac{2.4}{x+y}=\dfrac{4}{3}\end{matrix}\right.\\ Đặt.a=\dfrac{4}{x+y},b=\dfrac{4}{x-y}\\ \Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{5}{2}\\a+2b=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=\dfrac{5}{2}\\3a+6b=4\end{matrix}\right.\) 

\(\Leftrightarrow\left(3a+6b\right)-3a-3b=4-\dfrac{5}{2}\\ \Leftrightarrow3b=\dfrac{3}{2}\Rightarrow b=\dfrac{1}{2}\Rightarrow a+2.\dfrac{1}{2}=\dfrac{4}{3}\\\Leftrightarrow a+1=\dfrac{4}{3}\Rightarrow a=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{4}{x+y}=\dfrac{1}{3}\\\dfrac{4}{x+y}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=12\\x-y=8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\y=2\end{matrix}\right.\)

a/ Đơn giản là dùng phép thế:

\(x+2y+x+y+z=0\Rightarrow x+2y=0\Rightarrow x=-2y\)

\(x+y+z=0\Rightarrow z=-\left(x+y\right)=-\left(-2y+y\right)=y\)

Thế vào pt cuối:

\(\left(1-2y\right)^2+\left(y+2\right)^2+\left(y+3\right)^2=26\)

Vậy là xong

b/ Sử dụng hệ số bất định:

\(\left\{{}\begin{matrix}a\left(\frac{x}{3}+\frac{y}{12}-\frac{z}{4}\right)=a\\b\left(\frac{x}{10}+\frac{y}{5}+\frac{z}{3}\right)=b\end{matrix}\right.\)

\(\Rightarrow\left(\frac{a}{3}+\frac{b}{10}\right)x+\left(\frac{a}{12}+\frac{b}{5}\right)y+\left(\frac{-a}{4}+\frac{b}{3}\right)z=a+b\) (1)

Ta cần a;b sao cho \(\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}=-\frac{a}{4}+\frac{b}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}\\\frac{a}{3}+\frac{b}{10}=-\frac{a}{4}+\frac{b}{3}\end{matrix}\right.\) \(\Rightarrow\frac{a}{2}=\frac{b}{5}\)

Chọn \(\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\) thay vào (1):

\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)

\(\left\{{}\begin{matrix}\dfrac{2x+1}{4}-\dfrac{y-2}{3}=\dfrac{1}{12}\\\dfrac{x+5}{2}-\dfrac{y+7}{3}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}+\dfrac{1}{4}-\dfrac{y}{3}+\dfrac{2}{3}=\dfrac{1}{12}\\\dfrac{x}{2}+\dfrac{5}{2}-\dfrac{y}{3}-\dfrac{7}{3}=-4\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{5}{6}\\\dfrac{x}{2}-\dfrac{y}{3}=-\dfrac{25}{6}\end{matrix}\right.\) (vô lý)

Vậy HPT vô nghiệm

Đặt \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=a\\y-\dfrac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}=a^2-2\\y^2+\dfrac{1}{y^2}=b^2+2\end{matrix}\right.\)hệ đã cho tương đương:

\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\end{matrix}\right.\) \(\Rightarrow a^2+\left(3-a\right)^2-5=0\Rightarrow a^2-3a+2=0\)

\(\Rightarrow\left[{}\begin{matrix}a=1;b=2\\a=2;b=1\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=1\\y-\dfrac{1}{y}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-x+1=0\left(vn\right)\\y^2-2y-1=0\end{matrix}\right.\) (loại)

TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y-\dfrac{1}{y}=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2-2x+1=0\\y^2-y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=\dfrac{1-\sqrt{5}}{2}\\y=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy hệ đã cho có 2 cặp nghiệm:

\(\left(x;y\right)=\left(1;\dfrac{1-\sqrt{5}}{2}\right);\left(1;\dfrac{1+\sqrt{5}}{2}\right)\)

Đặt \(a=x+\dfrac{1}{x}\Leftrightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow x^2+\dfrac{1}{x^2}=a^2-2\)

\(b=y-\dfrac{1}{y}\Leftrightarrow b^2=y^2+\dfrac{1}{y^2}-2\Leftrightarrow y^2+\dfrac{1}{y^2}=b^2+2\)

Nên \(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}=5\Leftrightarrow a^2-2+b^2+2=5\Leftrightarrow a^2+b^2=5\)Vậy ta có hệ phương trình \(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\left(1\right)\end{matrix}\right.\)

Ta có a+b=3\(\Leftrightarrow b=3-a\)

Thay b=3-a vào (1)\(\Leftrightarrow a^2+\left(3-a\right)^2=5\Leftrightarrow a^2+9-6a+a^2=5\Leftrightarrow2a^2-6a+4=0\Leftrightarrow2\left(a^2-3a+2\right)=0\Leftrightarrow a^2-3a+2=0\Leftrightarrow a^2-a-2a+2=0\Leftrightarrow a\left(a-1\right)-2\left(a-1\right)=0\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-1=0\\a-2=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}b=2\\b=1\end{matrix}\right.\)

TH1:\(\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{1}{x}=1\\y-\dfrac{1}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-x+1=0\\y^2-2y-1=0\end{matrix}\right.\)

Ta có \(x^2-x+1=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy phương trình (2) vô nghiệm

TH2: \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y-\dfrac{1}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x^2-2x+1=0\\y^2-y-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=\dfrac{1\pm\sqrt{5}}{2}\end{matrix}\right.\)

Vậy (x,y)={(\(1;\dfrac{1+\sqrt{5}}{2}\));(\(1;\dfrac{1-\sqrt{5}}{2}\))}

Bài 2:

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

=>-4x-2y=3 và 8x+2y=-2

=>x=1/4; y=-2

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

=>y=6 và x-2=5/4

=>x=13/4; y=6

c: =>x+y=24 và 3x+y=78

=>-2x=-54 và x+y=24

=>x=27; y=-3

d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)

=>y+2=1 và x-1=25

=>x=26; y=-1