\(\sqrt{x^2+4x+3}+\sqrt{x^2+x}=\sqrt{3x^2+4x+1}\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}+\sqrt{x\left(x+1\right)}=\sqrt{\left(x+1\right)\left(3x+1\right)}\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x+3\right)}+\sqrt{x\left(x+1\right)}-\sqrt{\left(x+1\right)\left(3x+1\right)}=0\)

\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+3}+\sqrt{x}-\sqrt{3x+1}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x+1}=0\\\sqrt{x+3}+\sqrt{x}=\sqrt{3x+1}\end{cases}}\)

Suy ra x=-1 pt còn lại bình lên là thấy vô nghiệm

điều kiện xy \(\ge\) 0

\(\left\{{}\begin{matrix}x+y-\sqrt{xy}=7\\x^2+y^2+xy=133\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}\left(x+y\right)-\sqrt{xy}=7\\\left(x+y\right)^2-xy=133\end{matrix}\right.\)

đặc x + y = a ; xy = b

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-\sqrt{b}=7\\a^2-b=133\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7+\sqrt{b}\\\left(7+\sqrt{b}\right)^2-b=133\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7+\sqrt{b}\\49+14\sqrt{b}+b-b=133\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7+\sqrt{b}\\14\sqrt{b}=84\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=7+\sqrt{b}\\\sqrt{b}=6\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=13\\b=36\end{matrix}\right.\)

\(\Rightarrow\) x + y = 13 ; xy = 36

\(\Rightarrow\) x ; y là nghiệm của phương trình : x² - 13x + 36 = 0

bấm máy ta có : x = 4 ; x = 9

vậy x = 4 ; y = 9 hoặc x = 9 ; y = 4

xy = 36 (tmđk)

ĐKXĐ: \(\begin{cases}x+3\ge0\\ 3-2x\ge0\end{cases}\Rightarrow\begin{cases}x\ge-3\\ 2x\le3\end{cases}\Rightarrow-3\le x\le\frac32\)

\(x+4\sqrt{x+3}+2\sqrt{3-2x}=11\)

=>\(x-1+4\sqrt{x+3}-8+2\sqrt{3-2x}-2=0\)

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

=>\(x-1+4\cdot\frac{x+3-4}{\sqrt{x+3}+2}+2\cdot\frac{3-2x-1}{\sqrt{3-2x}+1}=0\)

=>\(x-1+4\cdot\frac{x-1}{\sqrt{x+3}+2}+2\cdot\frac{-2x+2}{\sqrt{3-2x}+1}=0\)

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

=>x-1=0

=>x=1(nhận)