Question

[Mo beo]

Giúp em vs 1,2 giải denta nha🙁

Answer 1

[Nguyễn Ngọc Huy Toàn]

a.\(5x^2-4x-1=0\)

\(\Delta=\left(-4\right)^2-4.5.\left(-1\right)=16+20=36>0\)

=> pt có 2 nghiệm phân biệt

\(\left\{{}\begin{matrix}x_1=\dfrac{4+\sqrt{36}}{10}=1\\x_2=\dfrac{4-\sqrt{36}}{10}=-\dfrac{1}{5}\end{matrix}\right.\)

b.\(7x^2-2\sqrt{7}x+1=0\)

\(\Delta=\left(-2\sqrt{7}\right)^2-4.1.7\)

    \(=28-28=0\)

=> pt có nghiệm kép

\(x_1=x_2=\dfrac{2\sqrt{7}}{2.7}=\dfrac{\sqrt{7}}{7}\)

Answer 2

[Vô danh]

a, \(\Delta=\left(-4\right)^2-4.5.\left(-1\right)=16+20=36>0\)

\(x_1=\dfrac{4+6}{10}=1\\ x_2=\dfrac{4-6}{10}=\dfrac{-1}{5}\)

\(b,\Delta=\left(-2\sqrt{7}\right)^2-4.7.1=28-28=0\)

\(\Rightarrow\) pt có nghiệm kép là: \(\dfrac{2\sqrt{7}}{14}=\dfrac{\sqrt{7}}{7}\)

\(c,4x^2-2=0\\ \Leftrightarrow2x^2-1=0\\ \Leftrightarrow x^2=\dfrac{1}{2}\\ \Leftrightarrow x=\pm\dfrac{\sqrt{2}}{2}\)

\(d,\left\{{}\begin{matrix}4x+y=3\\2x-3y=8\end{matrix}\right.\\ \Leftrightarrow \left\{{}\begin{matrix}y=3-4x\\2x-3\left(3-4x\right)=8\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}y=3-4x\\2x-9+12x=8\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}y=3-4x\\14x=17\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}y=3-4.\dfrac{17}{14}\\x=\dfrac{17}{14}\end{matrix}\right.\Leftrightarrow\\ \left\{{}\begin{matrix}y=-\dfrac{13}{7}\\x=\dfrac{17}{14}\end{matrix}\right.\)