Question

[Sách Giáo Khoa]

Cho các hàm số \(f\left(x\right)=x^2+2+\sqrt{2-x};g\left(x\right)=-2x^3-3x+5\)

                           \(u\left(x\right)=\left\{{}\begin{matrix}\sqrt{3-x};\left(x< 2\right)\\\sqrt{x^2-4};\left(x\ge2\right)\end{matrix}\right.\)

                          \(v\left(x\right)=\left\{{}\begin{matrix}\sqrt{6-x};\left(x\le0\right)\\x^2+1;\left(x>0\right)\end{matrix}\right.\)

Tính các giá trị \(f\left(-2\right)-f\left(1\right);f\left(-7\right)-g\left(-7\right);f\left(-1\right)-u\left(-1\right);u\left(3\right)-v\left(3;\right)v\left(0\right)-g\left(0\right);\dfrac{f\left(2\right)-f\left(-2\right)}{v\left(2\right)-v\left(-3\right)}\) ?

Answer 1

[Bùi Thị Vân]

\(f\left(-2\right)-f\left(1\right)=\left(-2\right)^2+2+\sqrt{2-\left(-2\right)}-\left(1^2+2+\sqrt{2-1}\right)\) \(=8-4=4\).
\(f\left(-7\right)-g\left(-7\right)=\left(-7\right)^2+2+\sqrt{2-\left(-7\right)}-\left(-2.\left(-7\right)^3-3.\left(-7\right)+5\right)=-658\)