Cho các hàm số
\(f(x) = {x^2} + 2 + \sqrt {2 - x} ;g(x) = - 2{x^3} - 3x + 5\);
\(u(x) = \left\{ \matrix{
\sqrt {3 - x} ,x < 2 \hfill \cr
\sqrt {{x^2} - 4} ,x \ge 2 \hfill \cr} \right.\);
\(v(x) = \left\{ \matrix{
\sqrt {6 - x} ,x \le 0 \hfill \cr
{x^2} + 1,x > 0 \hfill \cr} \right.\)
Tính các giá trị
\(f( - 2) - f(1);g(3);f( - 7) - g( - 7);f( - 1) - u( - 1);u(3) - v(3);v(0) - g(0);{{f(2) - f( - 2)} \over {v(2) - v( - 3)}}\)
Gợi ý làm bài
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\(f( - 2) - f( - 1) = {( - 2)^2} + 2 + \sqrt {2 + 2} - ({1^2} + 2 + \sqrt {2 - 1} ) = 8 - 4 = 4\);
\(g(3) = - {2.3^3} - 3.3 + 5 = - 58\);
\(f( - 7) - g( - 7) = {( - 7)^2} + 2 + \sqrt {2 + 7} - {\rm{[}} - 2.{( - 7)^3} - 3.( - 7) + 5] = - 658\);
\(f( - 1) - u( - 1) = 3 + \sqrt 3 - 2 = 1 + \sqrt 3 \);
\(u(3) - v(3) = \sqrt {9 - 4} - (9 + 1) = \sqrt 5 - 10\);
\(v(0) - g(0) = \sqrt 6 - 5\);
\({{f(2) - f( - 2)} \over {v(2) - v( - 3)}} = {{6 - 8} \over {5 - 3}} = - 1\)