Câu a ) 

\(ĐKXĐx\ne-1,3\)

Ta có : 

\(\frac{x}{2x+2}-\frac{2x}{x^2-2x-3}=\frac{x}{6-2x}\)

\(\Rightarrow\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=\frac{x}{-2\left(x-3\right)}\)

\(\Rightarrow\frac{x}{2\left(x+1\right)}.2\left(x+1\right)\left(x-3\right)-\frac{2x}{\left(x+1\right)\left(x-3\right)}.2\left(x+1\right)\left(x-3\right)\)

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

=> x(x-3) -4x =−x(x+1)

=> \(x^2-7x=-x^2-x\)

\(\Rightarrow2x^2-6x=0\)

\(\Rightarrow2x\left(x-3\right)=0\)

\(\Rightarrow x\in\left\{3,0\right\}\)

Câu b ) 

Ta có : 

\(\hept{\begin{cases}\sqrt{2}x-3y=2006\\2x+\sqrt{3}y=2007\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2}x-3y=2006\\2\sqrt{3}x+3y=2007\sqrt{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2}x-3y=2006\\2\sqrt{3}x+3y+\sqrt{2}x-3y=2007\sqrt{3}+2006\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{2}x-3y=2006\\\left(\sqrt{2}+2\sqrt{3}\right)x=2007\sqrt{3}+2006\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y=\frac{\sqrt{2}x-2006}{3}\\x=\frac{2007\sqrt{3}+2006}{\sqrt{2}+2\sqrt{3}}\end{cases}}\)

\(\hept{\begin{cases}y=\frac{\sqrt{2}.\frac{2007\sqrt{3}+2006}{\sqrt{2}+2\sqrt{3}}-2006}{3}\\x=\frac{2007\sqrt{3}+2006}{\sqrt{2}+2\sqrt{3}}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}y=\frac{2007\sqrt{6}-4012\sqrt{3}}{\left(\sqrt{2}+2\sqrt{3}\right).3}\\x=\frac{2007\sqrt{3}+2006}{\sqrt{2}+2\sqrt{3}}\end{cases}}\)

a) \(\hept{\begin{cases}\sqrt{2x}-\sqrt{3y}=1\left(1\right)\\x+\sqrt{3y}=\sqrt{2}\left(2\right)\end{cases}}\) ( ĐK \(x,y\ge0\) )

Từ (1) và (2)\(\Leftrightarrow\sqrt{2x}+x=1+\sqrt{2}\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\\sqrt{x}+\sqrt{2}+1=0\end{cases}}\)

\(\Leftrightarrow x=1\) ( Do \(x\ge0\) )

Thay \(x=1\) vào hệ (1) ta có :

\(\sqrt{2}-\sqrt{3y}=1\)

\(\Leftrightarrow\sqrt{3y}=\sqrt{2}-1\)

\(\Leftrightarrow y=\frac{3-2\sqrt{2}}{3}\) ( thỏa mãn )

P/s : E chưa học cái này nên không chắc lắm ...

\(b,\hept{\begin{cases}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\\left(\sqrt{2}-1\right)x+\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)y=\sqrt{2}-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\\left(\sqrt{2}-1\right)x+y=\sqrt{2}-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{2}-1\right)x-y=\sqrt{2}\\2y=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=\frac{\sqrt{2}-0.5}{\sqrt{2}-1}=\frac{3+\sqrt{2}}{2}\end{cases}}\)

\(d,\hept{\begin{cases}\sqrt{6x}-\sqrt{4y}=\sqrt{2}\\\sqrt{6x}+\sqrt{9y}=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}5\sqrt{y}=3-\sqrt{2}\\\sqrt{2x}+\sqrt{3y}=\sqrt{3}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{11-6\sqrt{2}}{25}\\x=\frac{9+6\sqrt{2}}{25}\end{cases}}\)

a ) \(HPT\Leftrightarrow\hept{\begin{cases}5x-y=4\left(1\right)\\3x-y=5\left(2\right)\end{cases}}\)

Lấy (1) trừ (2) :

\(\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\)

Thay \(x=-\frac{1}{2}\) vào (1) : \(y=5x-4=5.-\frac{1}{2}-4=-\frac{13}{2}\)

Vậy HPT có nghiệm \(\left(x,y\right)=\left(-\frac{1}{2},-\frac{13}{2}\right)\)

b ) \(\hept{\begin{cases}\sqrt{3}x-\sqrt{2}y=1\\\sqrt{2}x+\sqrt{3}y=\sqrt{3}\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{6}x-2y=\sqrt{2}\left(1\right)\\\sqrt{6}x+3y=3\left(2\right)\end{cases}}}\)

Lấy (2 ) -(1) thu được :

\(5y=3-\sqrt{2}\Rightarrow y=\frac{3-\sqrt{2}}{5}\)

Thay giá trị y trên vào (1) : \(x=\frac{2y+\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{6}+\sqrt{3}}{5}\)

Vậy ......

Em học lớp 4 thôi nên ko hiểu gì đâu ạ

\(\hept{\begin{cases}x-y=3\\\left(x-y\right).\left(x^2+xy+y^2\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=3\\x^2+xy+y^2=3\end{cases}\Leftrightarrow\hept{\begin{cases}y=x-3\\x^2+x.\left(x-3\right)+\left(x-3\right)^2=3\left(I\right)\end{cases}}}\)

Phương trình (I) tương đương: \(x^2+x^2-3x+x^2-6x+9=3\Leftrightarrow3x^2-9x+6=0\Rightarrow x^2-3x+2=0\)

\(\Leftrightarrow\left(x-1\right).\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}y=-2\\y=-1\end{cases}}}\)

Vậy \(\left(x,y\right)=\left(1,-2\right),\left(2,-1\right)\)

ôi thôi xong, e chịu

Câu 1: ĐK: x khác -1/2, y khác -2

Đặt \(\sqrt[3]{\frac{2x+1}{y+2}}=t\) Từ phương trình thứ nhất ta có:

\(t+\frac{1}{t}=2\Leftrightarrow t^2-2t+1=0\Leftrightarrow t=1\)

=> \(\sqrt[3]{\frac{2x+1}{y+2}}=1\Leftrightarrow2x+1=y+2\Leftrightarrow2x-y=1\)

Vậy nên ta có hệ phương trình cơ bản: \(\hept{\begin{cases}2x-y=1\\4x+3y=7\end{cases}}\)Em làm tiếp nhé>

\(1,ĐKXĐ:\hept{\begin{cases}y\ne-2\\x\ne-\frac{1}{2}\end{cases}}\)

Đặt \(\sqrt[3]{\frac{2x+1}{y+2}}=a\left(a\ne0\right)\)

\(Pt\left(1\right)\Leftrightarrow a+\frac{1}{a}=2\)

             \(\Leftrightarrow a^2+1=2a\)

             \(\Leftrightarrow\left(a-1\right)^2=0\)

            \(\Leftrightarrow a=1\)

           \(\Leftrightarrow\sqrt[3]{\frac{2x+1}{y+2}}=1\)

cảm ơn mọi người ạ <3

1) 

\(\hept{\begin{cases}\left(\sqrt{2}+\sqrt{3}\right)x-y\sqrt{2}=\sqrt{2}\\\left(\sqrt{2}+\sqrt{3}\right)x+y\sqrt{3}=-\sqrt{3}\end{cases}\Leftrightarrow\hept{\begin{cases}-y\left(\sqrt{2}+\sqrt{3}\right)=\sqrt{2}+\sqrt{3}\\\left(\sqrt{2}+\sqrt{3}\right)x+y\sqrt{3}=-\sqrt{3}\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)