Question

Câu 1: CHo hàm số : y = f(x) = 2x² - 10

a) Tính f (0) ;  f (1) ; f \(\left(-1\frac{1}{2}\right)\)

b) Tìm x biết f (x) = -2

Answer 1

a) Ta có : \(f\left(0\right)=2.0^2-10=-10\)

               \(f\left(1\right)=2.1^2-10=-8\)

               \(f\left(-1\frac{1}{2}\right)=f\left(\frac{-3}{2}\right)=2.\left(\frac{-3}{2}\right)^2-10=2.\frac{9}{4}-10=\frac{9}{2}-10=\frac{-11}{2}\)

b)Vì \(f\left(x\right)=2\)

\(\Rightarrow2x^2-10=-2\)

\(\Rightarrow2x^2=8\)

\(\Rightarrow x^2=4\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

Vậy \(x=2\)hoặc \(x=-2\)

Answer 2

a, \(f\left(0\right)=2.0^2-10=-10\)

\(f\left(1\right)=2.1^2-10=2-10=-8\)

Ta co \(-1\frac{1}{2}=-\frac{3}{2}\)

\(f\left(-\frac{3}{2}\right)=2.\left(-\frac{3}{2}\right)^2-10=2.\frac{9}{4}-10=\frac{18}{4}-\frac{40}{4}=-\frac{22}{4}=-\frac{11}{2}\)

b, Ta co : \(f\left(x\right)=-2\)hay \(2x^2-10=-2\Leftrightarrow2x^2=8\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)