\(2x^2+3x-5=0\)
\(< =>2x^2-2x+5x-5=0\)
\(< =>2x\left(x-1\right)+5\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(2x+5\right)=0\)
\(< =>\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
\(\hept{\begin{cases}x+2y=1\\-3x+4y=-18\end{cases}}\)
\(< =>\hept{\begin{cases}-3x-6y=-3\\-3x-6y+10y=-18\end{cases}}\)
\(< =>\hept{\begin{cases}x+2y=1\\10y=-18+3=-15\end{cases}}\)
\(< =>\hept{\begin{cases}x+2y=1\\y=-\frac{3}{2}\end{cases}< =>\hept{\begin{cases}x-3=1\\y=-\frac{3}{2}\end{cases}< =>\hept{\begin{cases}x=4\\y=-\frac{3}{2}\end{cases}}}}\)
Bài 1 : Ta có : \(\Delta=9-4\left(-5\right).2=9+40=49>0\)
\(x_1=\frac{-3-7}{4}=-\frac{11}{4};x_2=\frac{-3+7}{4}=1\)
Bài 2 :
\(\hept{\begin{cases}x+2y=1\\-3x+4y=-18\end{cases}\Leftrightarrow\hept{\begin{cases}2x+4y=2\\-3x+4y=-18\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}5x=20\\x+2y=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=-\frac{3}{2}\end{cases}}}\)
Vậy hệ pt có một nghiệm ( x ; y ) = ( 4 ; -3/2 )
1, Ta thấy : \(2+3+\left(-5\right)=0\)
Nên pt có nghiệm là \(\orbr{\begin{cases}x=1\\x=\frac{c}{a}=-\frac{5}{2}\end{cases}}\)
Bài 3 :
Với \(x>0\)
\(P=\left(\frac{1}{x+\sqrt{x}}-\frac{1}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+2\sqrt{x}+1}\)
\(=\left(\frac{1-\sqrt{x}}{x+\sqrt{x}}\right):\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)^2}=\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{x}=\frac{-\left(x-1\right)}{x}=\frac{-x+1}{x}\)