Question

giúp e vs ạ huhu

Answer 1

c/tiếp tục áp dụng công thức bậc 2 :

(a=12;b=-25;c=12) có:

\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2.a}\)

\(\Leftrightarrow x=\dfrac{-1.-25\pm\sqrt{-25^2-4.12.12}}{2.12}\)

\(\Leftrightarrow x=\dfrac{25\pm\sqrt{49}}{24}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{25+7}{24}\\x_2=\dfrac{25-7}{24}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{4}{3}\\x_2=\dfrac{3}{4}\end{matrix}\right.\)

từ trên suy ra:

\(\dfrac{3}{4}\le x\le\dfrac{4}{3}\)

 

Answer 2

b/áp dụng công thức bậc 2 :

\(x=\dfrac{-1.-3\pm\sqrt{3^2-4.2.-2}}{2.2}\)

\(\Leftrightarrow x=\dfrac{3\pm\sqrt{25}}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{3-5}{4}\\x_2=\dfrac{3+5}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x_1=-\dfrac{1}{2}\\x_2=2\end{matrix}\right.\)

Trên pc cj vẽ khó qué e tự nghiên cứu hỏi lại thầy cô nhe:<

\(\Rightarrow x\le-\dfrac{1}{2};x\ge2\)

\(\Rightarrow x\in\left\{-\infty;-\dfrac{1}{2}\right\}U\left\{\infty;2\right\}\)