Ko ai bt thì tôi tự giải. Xem có đúng ko?

Giải: 

Đặt: 

\(\hept{\begin{cases}a=x-1\\b=y-1\end{cases}}\)

Thay thế vào hệ, ta có:

\(\hept{\begin{cases}a+\sqrt{a^2+1}=3^b\\b+\sqrt{b^2+1}=3^a\end{cases}}\)

Vế trừ vế ta có:

\(a+\sqrt{a^2+1}+3^a=b+\sqrt{a^2+1}+3^b\)

Dùng hàm số 

Suy ra: \(a=b\)

a=b nha anh k em nha

có thể bn nhưng lớp mk chưa học đến dạng này

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

\(\Leftrightarrow\left(x^3+y^3\right)^2=\left(x^2+y^2\right)^2.\left(x^2-xy+y^2\right)\)

\(\Leftrightarrow\left(x+y\right)^2.\left(x^2-xy+y^2\right)^2=\left(x^2+y^2\right)^2.\left(x^2-xy+y^2\right)\)

\(\Leftrightarrow\left(x+y\right)^2.\left(x^2-xy+y^2\right)=\left(x^2+y^2\right)^2\)

\(\Leftrightarrow\left(x^3+y^3\right)\left(x+y\right)=\left(x^2+y^2\right)^2\)

\(\Leftrightarrow x^4+x^3y+xy^3+y^4=x^4+y^4+2x^2y^2\)

\(\Leftrightarrow x^3y+xy^3-2x^2y^2=0\)

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

\(\Leftrightarrow\sqrt{4x-3}.\left(x-y\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{4x-3}=0\\\left(x-y\right)^2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}4x-3=0\\x-y=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{4}\\x=y\end{cases}}\)

Xét trường hợp:

Với x=3/4

=>\(x=\frac{3}{4}\Leftrightarrow y.\frac{3}{4}=0\Leftrightarrow y=0\)

Với: \(x=y\)

Có: \(xy=\sqrt{4x-3}\Leftrightarrow x^2y^2=4x-3\Leftrightarrow x^4-4x+3=0\Leftrightarrow x\left(x^3-1\right)-3\left(x-1\right)=0\)\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)\left(x^2+2x+3\right)=0\)( vì x^2+2x+3 luôn dương. Tự c/m nhé )

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)\(\Leftrightarrow x=y=1\)

KL:.................................

thanks anh ạ 

\(\hept{\begin{cases}x+\sqrt{x^2+1}=y+\sqrt{y^2-1}\left(1\right)\\x^2+y^2-xy=1\left(2\right)\end{cases}}\) \(DK:y^2\ge1\) 

Đặt: \(x^2+\sqrt{x^2+1}=y^2+\sqrt{y^2-1}=t\) . Vì: \(\sqrt{x^2+1}>\sqrt{x^2}=\left|x\right|\ge-x\Rightarrow t=\sqrt{x^2+1}+x>0\)  

\(\Leftrightarrow\hept{\begin{cases}x-t=\sqrt{x^2+1}\\y-t=\sqrt{y^2-1}\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2-2xt+t^2=x^2+1\\y^2-2yt+t^2=y^2-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{t^2-1}{2t}\\y=\frac{t^2+1}{2t}\end{cases}}\) 

\(\left(2\right)\Rightarrow\left(\frac{t^2-1}{2t}\right)^2+\left(\frac{t^2+1}{2t}\right)^2-\left(\frac{t^2-1}{2t}\right)\left(\frac{t^2+1}{2t}\right)=1\) 

\(\Leftrightarrow\frac{t^4-2t^2+1+t^4+2t^2+1-t^4+1}{4t^2}=1\) 

\(\Leftrightarrow t^4+3=4t^2\Leftrightarrow t^4-4t^2+3=0\Leftrightarrow\orbr{\begin{cases}t^2=1\\t^2=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}t=1\\t=\sqrt{3}\end{cases}}\) 

Với \(t=1\Rightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}\left(TM\right)\)

Với \(t=\sqrt{3}\Rightarrow\hept{\begin{cases}x=\frac{1}{\sqrt{3}}\\y=\frac{2}{\sqrt{3}}\end{cases}}\left(TM\right)\) 

Vậy:....

Xét pt (1) ta có

PT (1) <=> x - y = \(\sqrt{y^2-1}-\sqrt{x^2+1}\)

<=> xy = \(1\sqrt{\left(x^2+1\right)\left(y^2-1\right)}\)

<=> y² - x² = 1

Thế vào pt (2) ta được

y² + x² - xy = y² - x²

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

Tới đây thì đơn giản rồi

\(\hept{\begin{cases}\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)=2012\left(1\right)\\x^2+z^2-4\left(y+z\right)+8=0\left(2\right)\end{cases}}\)

Ta có:(1) \(\Leftrightarrow\left(x+\sqrt{x^2+2012}\right)\left(y+\sqrt{y^2+2012}\right)\left(\sqrt{y^2+2012}-y\right)\)\(=2012\left(\sqrt{y^2+2012}-y\right)\)(Do \(\sqrt{y^2+2012}-y\ne0\forall y\))

\(\Leftrightarrow2012\left(x+\sqrt{x^2+2012}\right)=2012\left(\sqrt{y^2+2012}-y\right)\)

\(\Leftrightarrow x+\sqrt{x^2+2012}=\sqrt{y^2+2012}-y\)\(\Leftrightarrow x+y=\sqrt{y^2+2012}-\sqrt{x^2+2012}\)

\(\Leftrightarrow x+y=\)\(\frac{\left(\sqrt{y^2+2012}+\sqrt{x^2+2012}\right)\left(\sqrt{y^2+2012}-\sqrt{x^2+2012}\right)}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}\)

\(\Leftrightarrow x+y=\frac{y^2-x^2}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}\)\(\Leftrightarrow\left(x+y\right)\frac{\sqrt{y^2+2012}-y+\sqrt{x^2+2012}+x}{\sqrt{y^2+2012}+\sqrt{x^2+2012}}=0\)

Do \(\hept{\begin{cases}\sqrt{y^2+2012}>\sqrt{y^2}=\left|y\right|\ge y\forall y\\\sqrt{x^2+2012}>\sqrt{x^2}=\left|x\right|\ge-x\forall x\end{cases}}\)\(\Rightarrow\sqrt{y^2+2012}-y+\sqrt{x^2+2012}+x>0\forall x,y\Rightarrow x+y=0\)

\(\Rightarrow y=-x\)

Thay y = -x vào (2), ta được: \(x^2+z^2+4x-4z+8=0\)

\(\Leftrightarrow\left(x+2\right)^2+\left(z-2\right)^2=0\Leftrightarrow\hept{\begin{cases}x=-2\\z=2\end{cases}}\Rightarrow y=-x=2\)

Vậy hệ có nghiệm \(\left(x;y;z\right)=\left(-2;2;2\right)\)

Ôi trời nhiều thía ? làm từng câu một ha !

a \(\hept{\begin{cases}\left(x+5\right)\left(y-2\right)=\left(x+2\right)\left(y-1\right)\\\left(x-4\right)\left(y+7\right)=\left(x-3\right)\left(y+4\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}xy-2x+5y-10=xy-x+2y-2\\xy+7x-4y-28=xy+4x-3y-12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-x+3y=8\\3x-y=16\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y=16\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\3x-y-3x+9y=16+24\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-3x+9y=24\\8y=40\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=7\\y=5\end{cases}}\)

b, ĐKXĐ \(x\ne\pm y\)

Đặt \(\frac{1}{x+y}=a\)  và  \(\frac{1}{x-y}=b\)(a và b khác 0)

Ta có hệ \(\hept{\begin{cases}a-2b=2\\5a-4b=3\end{cases}}\)

          \(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b=3\end{cases}}\)

       \(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\5a-4b-2a+4b=3-4\end{cases}}\)

       \(\Leftrightarrow\hept{\begin{cases}2a-4b=4\\3a=-1\end{cases}}\)

      \(\Leftrightarrow\hept{\begin{cases}a=-\frac{1}{3}\\b=-\frac{7}{6}\end{cases}}\)

    \(\Leftrightarrow\hept{\begin{cases}\frac{1}{x+y}=-\frac{1}{3}\\\frac{1}{x-y}=-\frac{7}{6}\end{cases}}\)

   \(\Leftrightarrow\hept{\begin{cases}x+y=-3\\x-y=-\frac{6}{7}\end{cases}}\)

  \(\Leftrightarrow\hept{\begin{cases}x+y-x+y=-3+\frac{6}{7}\\x-y=-\frac{6}{7}\end{cases}}\)

  \(\Leftrightarrow\hept{\begin{cases}2y=-\frac{15}{7}\\x-y=-\frac{6}{7}\end{cases}}\)

  \(\Leftrightarrow\hept{\begin{cases}x=-\frac{27}{14}\\y=-\frac{15}{14}\end{cases}}\)

c,\(\hept{\begin{cases}4x^2+y^2=13\\2x^2-y^2=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}4x^2+y^2+2x^2-y^2=13-7\\2x^2-y^2=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}6x^2=6\\2x^2-y^2=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=9\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\pm1\\y=\pm3\end{cases}}\)

t ms lên 7  =>  I don't know