mong mọi người giải giúp em vs 

\(\frac{\left(x+4\right)\left(x-2\right)}{x^2-2x+3}=\left(x+1\right)\frac{x+2-4}{\sqrt{x+2}+2}\)

\(\left(x-2\right)\left(\frac{x+4}{x^2-2x+3}-\frac{x+1}{\sqrt{x+2}+2}\right)=0\)

+ x=2

+ chiu kho lam cai con lai

Mk nghĩ đề là như này : \(2\left(x^2+2\right)=3\left(\sqrt{x^3+8}+2x\right)\)

fix lai chut...

...

Ta có : \(a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-2x+4=4x+8\)

\(\Leftrightarrow x^2-6x-4=0\)

\(\Delta=6^2-4\cdot\left(-4\right)=52\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{6+\sqrt{52}}{2}=3+\sqrt{13}\\x=\frac{6-\sqrt{52}}{2}=3-\sqrt{13}\end{matrix}\right.\)

Vậy....

ĐK: \(x\ge-2\)

\(2\left(x^2+2\right)=3\left(\sqrt{x^3+8}+2x\right)\)

\(\Leftrightarrow2x^2+4=3\sqrt{x^3+8}+6x\)

\(\Leftrightarrow2x^2-6x+4=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow2\left(x^2-3x+2\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x+4}=a\\\sqrt{x+2}=b\end{matrix}\right.\)( \(a,b\ge0\) )

Ta có : \(a^2-b^2=x^2-2x+4-x-2=x^2-3x+2\)

\(pt\Leftrightarrow2\left(a^2-b^2\right)=3ab\)

\(\Leftrightarrow2a^2-3ab-2b^2=0\)

\(\Leftrightarrow\left(a-2b\right)\left(2a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\left(chon\right)\\2a=-b\left(loai\right)\end{matrix}\right.\)

Ta có \(a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-4x+4=4x+8\)

\(\Leftrightarrow x^2-8x-4=0\)

\(\Delta=8^2-4\cdot\left(-4\right)=80\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{8+\sqrt{80}}{2}\\x=\frac{8-\sqrt{80}}{2}\end{matrix}\right.\)( thỏa )

Vậy...

\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3-3xy\left(x+y\right)+\left(xy\right)^3+7\left(xy+x+y+1\right)=31\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)=2\\\left(x+y\right)^3+\left(xy\right)^3+7\left(xy+x+y\right)=30\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) với \(u^2\ge4v\)

\(\Rightarrow\left\{{}\begin{matrix}uv=2\\u^3+v^3+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3-3uv\left(u+v\right)+7\left(u+v\right)=30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\\left(u+v\right)^3+\left(u+v\right)-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=2\\u+v=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=2\\v=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=2\\xy=1\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(1;1\right)\)

2.

ĐKXĐ: \(0\le x\le\dfrac{3}{2}\)

\(\Leftrightarrow9x\left(3-2x\right)+81+54\sqrt{x\left(3-2x\right)}=49x+25\left(3-2x\right)+70\sqrt{x\left(3-2x\right)}\)

\(\Leftrightarrow9x^2-14x-3+8\sqrt{x\left(3-2x\right)}=0\)

\(\Leftrightarrow9\left(x^2-2x+1\right)-4\left(3-x-2\sqrt{x\left(3-2x\right)}\right)=0\)

\(\Leftrightarrow9\left(x-1\right)^2-\dfrac{36\left(x-1\right)^2}{3-x+2\sqrt{x\left(3-2x\right)}}=0\)

\(\Leftrightarrow9\left(x-1\right)^2\left(1-\dfrac{4}{3-x+2\sqrt{x\left(3-2x\right)}}\right)=0\)

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

\(\left(1\right)\Leftrightarrow2\sqrt{x\left(3-2x\right)}=x+1\)

\(\Leftrightarrow4x\left(3-2x\right)=x^2+2x+1\)

\(\Leftrightarrow9x^2-10x+1=0\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{9}\end{matrix}\right.\)