\(\hept{\begin{cases}\left|x-2\right|+2\sqrt{y+3}=9\\x+\sqrt{y+3}=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left|x-2\right|+2\sqrt{y+3}=9\\x-2+\sqrt{y+3}=-3\end{cases}}\)(1)

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

Xét: \(x\ge2\)

=> (1) trở thành \(\Leftrightarrow\hept{\begin{cases}x-2+2\sqrt{y+3}=9\\x-2+\sqrt{y+3}=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a+2b=9\\a+b=-3\end{cases}}\)

Xét \(x< 2\)

=> (1) trở thành \(\Leftrightarrow\hept{\begin{cases}-\left(x-2\right)+2\sqrt{y+3}=9\\x-2+\sqrt{y+3}=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}-a+2b=9\\a+b=-3\end{cases}}\)

Từ hệ pt trên \(< =>\hept{\begin{cases}|x-2|+2\sqrt{y+3}=9\\x+\sqrt{y+3}=-1\end{cases}}\)

\(< =>\hept{\begin{cases}|x-2|+2\sqrt{y+3}=9\\2x+2\sqrt{y+3}=-2\end{cases}}\)

\(< =>\hept{\begin{cases}|x-2|-2x=11\\x+\sqrt{y+3}=-1\end{cases}}\)

Xét \(x\ge2\)=>  \(|x-2|=\left(x-2\right)\)

\(< =>\hept{\begin{cases}x-2-2x=11\\x+\sqrt{y+3}=-1\end{cases}}\)

\(< =>\hept{\begin{cases}x=-13\\-13+\sqrt{y+3}=-1\end{cases}}\)

\(< =>\hept{\begin{cases}x=-13\\\sqrt{y+3}=12\end{cases}}\)

\(< =>\hept{\begin{cases}x=-13\\\sqrt{y+3}=\sqrt{144}\end{cases}}\)

\(< =>\hept{\begin{cases}x=-13\\y=141\end{cases}}\)

Có ai check cái :( e mới học dạng này nên chưa chắc :(((

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

Ta có : 

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{4+\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}\)

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

\(=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1\)

bài này e chịu , nhưng sau những lần tìm kiếm thì đây ạ 

\(4x^2=3x+4\)

\(4x^2-3x-4=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{73}}{8}\\x=\frac{3-\sqrt{73}}{8}\end{cases}\Rightarrow\orbr{\begin{cases}x=1,443\\x=-0,693\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{73}}{8}\\x=\frac{3-\sqrt{73}}{8}\end{cases}}\)

Chỉ cần tính đến đó thôi e .... ạ :P

Học tốt!!!!!!!!

ĐK: \(\hept{\begin{cases}x\ge1\\x^2-20x+24\le0\end{cases}}\)

\(x^2-20x+24+8\sqrt{3\left(x-1\right)}=0\)

\(\Leftrightarrow2\left(x^2-20x+24+8\sqrt{3x-3}\right)=0\)

\(\Leftrightarrow2x^2-32x+32+8\left(2\sqrt{3x-3}-x+2\right)=0\)

\(\Leftrightarrow2x^2-32x+32+8\left[2\sqrt{3x-3}-\left(x-2\right)\right]=0\)

\(\Leftrightarrow2x^2-32x+32+8\frac{4\left(3x-3\right)-\left(x-2\right)^2}{2\sqrt{3x-3}+x-2}=0\)

\(\Leftrightarrow2x^2-32x+32+8\frac{12x-12-x^2+4x-4}{2\sqrt{3x-3}+x-2}=0\)

\(\Leftrightarrow2\left(x^2-16x+16\right)-8\frac{x^2-16x+16}{2\sqrt{3x-3}+x-2}=0\)

\(\Leftrightarrow\left(x^2-16x+16\right)\left(2-\frac{8}{2\sqrt{3x-3}+x-2}\right)=0\)

Xét \(2-\frac{8}{2\sqrt{3x-3}+x-2}=0\)

\(\Leftrightarrow2\sqrt{3x-3}+x-6=0\)

\(\Leftrightarrow\left(2\sqrt{3x-3}\right)^2=\left(6-x\right)^2\)

\(\Leftrightarrow12x-12=x^2-12x+36\)

\(\Leftrightarrow0=x^2-24x+48\)

Tự làm tiếp nhé ~

a, \(2\left(x+3\right)\left(x-4\right)=\left(2x-1\right)\left(x+2\right)-27\)

\(\Leftrightarrow2\left(x^2-4x+3x-12\right)=2x^2+4x-x-2-27\)

\(\Leftrightarrow2x^2-2x-24=2x^2+3x-29\Leftrightarrow-5x+5=0\Leftrightarrow x=1\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-3\right)\left(x+3\right)=26\)

\(\Leftrightarrow x^3-8-x\left(x^2-9\right)=26\Leftrightarrow-8+9x=26\)

\(\Leftrightarrow9x=18\Leftrightarrow x=2\)

<=> \(\hept{\begin{cases}7x-3y=5\\3x+2y=12\end{cases}}\) <=> \(\hept{\begin{cases}21x-9y=15\\21x+14y=84\end{cases}}\) <=> \(\hept{\begin{cases}-23y=-69\\3x+2y=12\end{cases}}\) <=> \(\hept{\begin{cases}y=3\\x=\frac{12-2y}{3}=\frac{12-2.3}{3}=2\end{cases}}\)

Vậy nghiệm của hpt là: (2;3)