Tính giới hạn sau: \(\mathop {\lim }\limits_{x \to 0} \frac{{\cos 3x - \cos 7x}}{{{x^2}}}\)

2x - 3

0

3 - 2x

2x + 3

\(\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} } \)

\(\int\limits_{ - a}^a {f(x)dx = -\int\limits_0^a {f(x)dx} } \)

\(\int\limits_{ - a}^a {f(x)dx = \int\limits_0^a {f(x)dx} } \)

\(\int\limits_{ - a}^a {f(x)dx = 2\int\limits_{ - a/2}^{a/2} {f(x)dx} } \)

\(\frac{1}{5}\)

\(\frac{1}{64}\)

\(\frac{1}{8}\)

\(\infty\)

\((\forall x \in \left[ {a,b} \right])f(x) < g(x) \Rightarrow \int\limits_a^b {f(x)dx} > \int\limits_a^b {g(x)dx} \)

\((\forall x \in \left[ {a,b} \right])f(x) \le g(x) \Rightarrow \int\limits_a^b {f(x)dx} \le \int\limits_a^b {g(x)dx} \)

\((\forall x \in \left[ {a,b} \right])f(x) \le g(x) \Rightarrow \int\limits_a^b {f(x)g(x)dx} \le \int\limits_a^b {g(x)dx} \)

\(f(x) \le g(x) \Rightarrow \int\limits_a^b {g(x)dx} \le \int\limits_a^b {g(x)dx} \)

\(\frac{1}{5}\)

0

\(\infty \)

\(\frac{1}{10}\)

\(1+ln2\)

\(1-ln2\)

\(\frac{1}{5}\ln 2\)

\(\frac{12}{5}\ln 6\)

\(- 2\ln \frac{3}{{4 + \sqrt 7 }}\)

0

\(\ln \frac{3}{{4 + \sqrt 7 }}\)

\( 2\ln \frac{3}{{4 + \sqrt 7 }}\)

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Chuỗi có dấu bất kỳ

Kết quả khác

r = 1/5

r = 3

r = 5

\(2ln2\)

\(1- 2ln2\)

\(1-ln2\)

\(2-2ln2\)