a) bạn thử bình phương 2 vế xem sao
b) tìm x ở pt 1 thế vào pt2
\(a,ĐKXĐ:x\ge-\frac{10}{3}\)
Ta có: \(x^2+9x+20=2\sqrt{3x+10}\)
\(\Leftrightarrow\left(x^2+6x+9\right)+\left(3x+10-2\sqrt{3x+10}+1\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2+\left(\sqrt{3x+10}-1\right)^2=0\)
Do \(VT\ge0\forall x\)
Nên dấu ''='' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+3=0\\\sqrt{3x+10}-1=0\end{cases}\Leftrightarrow x=-3}\)(Tm ĐKXĐ)
Vậy pt có nghiệm x = -3
\(b,\hept{\begin{cases}y-x=xy\\4x+2y=5xy\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y-x=xy\\4x+2y=5y-5x\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y-x=xy\\9x=2y\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y-x=xy\\x=\frac{2y}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y-\frac{2y}{9}=\frac{2y^2}{9}\\x=\frac{2y}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}9y-2y=2y^2\\x=\frac{2y}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2y^2-7y=0\\x=\frac{2y}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=0\\x=\frac{2y}{9}\end{cases}}\left(h\right)\hept{\begin{cases}y=\frac{7}{2}\\x=\frac{2y}{9}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=0\\x=0\end{cases}\left(h\right)\hept{\begin{cases}y=\frac{7}{2}\\x=\frac{7}{9}\end{cases}}}\)
Vậy .......
