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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
\(f( – 2) – f( – 1) = {( – 2)^2} + 2 + \sqrt {2 + 2} – ({1^2} + 2 + \sqrt {2 – 1} ) = 8 – 4 = 4\);
\(g(3) = – {2.3^3} – 3.3 + 5 = – 58\);
\(f( – 7) – g( – 7) = {( – 7)^2} + 2 + \sqrt {2 + 7} – {\rm{[}} – 2.{( – 7)^3} – 3.( – 7) + 5] = – 658\);
\(f( – 1) – u( – 1) = 3 + \sqrt 3 – 2 = 1 + \sqrt 3 \);
\(u(3) – v(3) = \sqrt {9 – 4} – (9 + 1) = \sqrt 5 – 10\);
\(v(0) – g(0) = \sqrt 6 – 5\);
\({{f(2) – f( – 2)} \over {v(2) – v( – 3)}} = {{6 – 8} \over {5 – 3}} = – 1\)