\(\Leftrightarrow\frac{y+x}{xy}=\frac{1}{2}\)

=>\(\frac{x+y}{xy}-\frac{1}{2}=0\)

\(\Rightarrow\frac{-\left(x-2\right)y-2x}{2xy}=0\)

=>(x-2)y-2x=0

=>x-2=0( vì x-2=0 thì nhân y-2x ms =0 )

=>x=2

=>y-2=0

=>y=2

vậy x=y=2

1. a = 3 thì phương trình trở thành:

\(\frac{x+3}{3-x}-\frac{x-3}{3+x}=\frac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2}-x^2\)

\(\Leftrightarrow\frac{\left(x+3\right)^2+\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\frac{-3\left[-9+1\right]}{9}-x^2\)

\(\Leftrightarrow\frac{x^2+6x+9+x^2-6x+9}{\left(3-x\right)\left(3+x\right)}=\frac{-3.\left(-8\right)}{9}-x^2\)

\(\Leftrightarrow\frac{2x^2+18}{9-x^2}=\frac{24}{9}-x^2\)

\(\Leftrightarrow\frac{2x^2+18}{9-x^2}+x^2=\frac{24}{9}\)

\(\Leftrightarrow\frac{2x^2+18+9x^2-x^4}{9-x^2}=\frac{24}{9}\)

\(\Leftrightarrow\frac{11x^2+18-x^4}{9-x^2}=\frac{24}{9}\)

\(\Leftrightarrow99x^2+18-9x^4=216-24x^2\)

\(\Leftrightarrow9x^4-123x^2+198=0\)

Đặt \(x^2=t\left(t\ge0\right)\)

Phương trình trở thành \(9t^2-123t+198=0\)

Ta có \(\Delta=123^2-4.9.198=8001,\sqrt{\Delta}=3\sqrt{889}\)

\(\Rightarrow\orbr{\begin{cases}t=\frac{123+3\sqrt{889}}{18}=\frac{41+\sqrt{889}}{6}\\t=\frac{123-3\sqrt{889}}{18}=\frac{41-\sqrt{889}}{6}\end{cases}}\)

Lúc đó \(\orbr{\begin{cases}x^2=\frac{41+\sqrt{889}}{6}\\x^2=\frac{41-\sqrt{889}}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm\sqrt{\frac{41+\sqrt{889}}{6}}\\x=\pm\sqrt{\frac{41-\sqrt{889}}{6}}\end{cases}}\)

Vậy pt có 4 nghiệm \(S=\left\{\pm\sqrt{\frac{41+\sqrt{889}}{6}};\pm\sqrt{\frac{41-\sqrt{889}}{6}}\right\}\)

Sửa)):

a = -3 mà ghi lôn a = 3.giải tương tự như 3

\(\left(1+x\sqrt{x^2+1}\right)\left(\sqrt{x^2+1}-x\right)=1\)

\(\Rightarrow\dfrac{1+x\sqrt{x^2+1}}{\sqrt{x^2+1}+x}=1\)

\(\Rightarrow1+x\sqrt{x^2+1}=\sqrt{x^2+1}+x\)

\(\Rightarrow1+x\sqrt{x^2+1}-\sqrt{x^2+1}-x=0\)

\(\Rightarrow-\left(x-1\right)+\left(x-1\right)\sqrt{x^2+1}=0\)

\(\Rightarrow\left(x-1\right)\left(\sqrt{x^2+1}-1\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\\sqrt{x^2+1}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x^2+1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

\(a,2y^2-x+2xy=y+4\\ \Leftrightarrow2y\left(x+y\right)-\left(x+y\right)=4\\ \Leftrightarrow\left(2y-1\right)\left(x+y\right)=4=4\cdot1=\left(-4\right)\left(-1\right)=\left(-2\right)\left(-2\right)=2\cdot2\)

Vì \(x,y\in Z\Leftrightarrow2y-1\) lẻ 

\(\left\{{}\begin{matrix}2y-1=1\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2y-1=-1\\x+y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=0\end{matrix}\right.\)

Vậy PT có nghiệm \(\left(x;y\right)=\left\{\left(3;1\right);\left(4;0\right)\right\}\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu