\(a,\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{2}{y}=2\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{5}{y}=3\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{5}{3}\\\dfrac{2}{x}+\dfrac{9}{5}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

\(b,\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}-\dfrac{135}{y}=525\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{9}{y}=35\\-\dfrac{163}{y}=489\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-27=35\\y=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{31}\\y=-\dfrac{1}{3}\end{matrix}\right.\)

a: Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\)

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

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

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

b, \(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}\\x+y=10\end{matrix}\right.\)Theo tc dãy tỉ số bằng nhau 

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{10}{5}=2\Rightarrow x=4;y=6\)

a.\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y=6\\2x-3y=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=15\\2x-3y=9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\2.3-3y=9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

b.\(\Leftrightarrow\left\{{}\begin{matrix}3x=2y\\x+y-10=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\x+y-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2y=0\\2x+2y=20\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=20\\3x-2y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=4\\3.4-2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

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

b, ĐKXĐ:\(y\ne0\)

\(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{10-y}{y}=\dfrac{2}{3}\\x=10-y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}3\left(10-y\right)=2y\\x=10-y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}30-3y=2y\\x=10-y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}5y=30\\x=10-y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=6\\x=10-6\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=6\\x=4\left(tm\right)\end{matrix}\right.\)

a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)

b: Ta có: \(\left\{{}\begin{matrix}\left(x+5\right)\left(y-4\right)=xy\\\left(x+5\right)\left(y+12\right)=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy-4x+5y-20-xy=0\\xy+12x+5y+60-xy=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-4x+5y=20\\12x+5y=-60\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-16y=80\\-4x+5y=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\-4x=20-5y=20-5\cdot\left(-5\right)=45\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-5\\x=-\dfrac{45}{4}\end{matrix}\right.\)

Câu 1: 

a) Ta có: \(x^4+3x^2-4=0\)

\(\Leftrightarrow x^4+4x^2-x^2-4=0\)

\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)

\(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)

mà \(x^2+4>0\forall x\)

nên \(x^2-1=0\)

\(\Leftrightarrow x^2=1\)

hay \(x\in\left\{1;-1\right\}\)

Vậy: S={1;-1}

Câu 1: 

b) Ta có: \(\left\{{}\begin{matrix}x+2y=5\\x-5y=-9\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}7y=14\\x+2y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=5-2y=1\end{matrix}\right.\)

Vậy: (x,y)=(1;2)

Câu 2: 

a) Thay x=-2 vào (d), ta được:

\(y=-2\cdot\left(-2\right)-6=4-6=-2\)

Vậy: T(-2;-2) thuộc (d)

\(a,\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\ b,\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\23y=46\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ d,\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

\(e,\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=10\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x=20\\6x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y=46\\5x+2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

e. \(\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\4x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a) \(\begin{cases} 3x -y=5\\ 4x +2y=10 \end{cases} \)

    \(\begin{cases} 12x - 4y= 20\\ 12x +6y= 30 \end{cases} \)

    \(\begin{cases} -10y=-10\\ 3x-y=5 \end{cases} \)

    \(\begin{cases} y=1\\ 3x-1=5 \end{cases} \)

    \(\begin{cases} y=1\\ 3x=6 \end{cases} \)

    \(\begin{cases} y=1\\ x=2 \end{cases} \)

Hpt có nghiệm duy nhất: {1;2}

b)\(\begin{cases} 5x +2y=9\\ x+5y=11 \end{cases} \)

  \(\begin{cases} 5x+2y=9\\ 5x+25y=55 \end{cases} \)

  \(\begin{cases} -23y=-46\\ x+5y=11 \end{cases} \)

  \(\begin{cases} y=2\\ x+ 5*2=11 \end{cases} \)

  \(\begin{cases} y=2\\ x+10=11 \end{cases} \)

Hpt có nghiệm duy nhất:{1;2}

c)\(\begin{cases} 3x+y=10\\ 4x-3y=9 \end{cases} \)

  \(\begin{cases} 12x+4y=40\\ 12x-9y=27 \end{cases} \)

 \(\begin{cases} 13y=13\\ 3x+y=10 \end{cases} \)

 \(\begin{cases} y=1\\ 3x+1=10 \end{cases} \)

 \(\begin{cases} y=1\\ 3x=9 \end{cases} \)

hpt có nghiệm duy nhất:{1;3}

d)\(\begin{cases} 4x+3y=22\\ 5x+3y=26 \end{cases} \)

  \(\begin{cases} 20x+15y=110\\ 20x+12y=104 \end{cases} \)

 \(\begin{cases} 3y=6\\ 4x+3y=22 \end{cases} \)

 \(\begin{cases} y=2\\ 4x+3*2=22 \end{cases} \)

 \(\begin{cases} y=2\\ 4x+6=22 \end{cases} \)

hệ phương trình có nghiệm duy nhất:{2;4}

e)\(\begin{cases} 4x-3y=5\\ 5x+3y=13 \end{cases} \)

  \(\begin{cases} 20x-15y=25\\ 20x+12y=52 \end{cases} \)

 \(\begin{cases} -27y=-27\\ 4x-3y=5 \end{cases} \)

  \(\begin{cases} y=1\\ 4x-3*1=5 \end{cases} \)

   \(\begin{cases} y=1\\ 4x-3=5 \end{cases} \)

Hệ phương trình có nghiệm duy nhất là:{1;2}

a: \(\sqrt{x^2+6x+9}=\sqrt{11+6\sqrt{2}}\)

=>\(\sqrt{\left(x+3\right)^2}=\sqrt{\left(3+\sqrt{2}\right)^2}\)

=>\(\left|x+3\right|=\left|3+\sqrt{2}\right|=3+\sqrt{2}\)

=>\(\left[{}\begin{matrix}x+3=3+\sqrt{2}\\x+3=-3-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-6-\sqrt{2}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}4x-2y=8\\x+2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y+x+2y=8-3\\2x-y=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5x=5\\y=2x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\cdot1-4=-2\end{matrix}\right.\)