Đề 3:
Câu 1:
a) Ta có: \(2x^2-3x-3=0\)
\(\Delta=\left(-3\right)^2-4\cdot2\cdot\left(-3\right)=9+24=33\)
Vì \(\Delta>0\) nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+\sqrt{33}}{2}\\x_2=\dfrac{3-\sqrt{33}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3+\sqrt{33}}{2};\dfrac{3-\sqrt{33}}{2}\right\}\)
(5x-3)(1+x)=21
\(\Leftrightarrow\) 5x+\(5x^2\)-3-3x=21
\(\Leftrightarrow\)\(5x^2\)+2x-24=0 (✳)
ta có: \(\Delta\)'= \(1^2\)-5\(\times\)(-24)=121>0
\(\sqrt{\Delta'}\)= \(\sqrt{121}\)=11
\(\Rightarrow\) pt (✳) có 2 nghiệm phân biệt
\(x_1\)=\(\dfrac{-1+11}{5}\)=2
\(x_2\)=\(\dfrac{-1-11}{5}\)=\(\dfrac{-12}{5}\)
vậy....
