Câu hỏi:

Ta có: $\displaystyle\int\limits_{0}^{m}(3x^2-2x+1)dx = x^3 - x^2 + x \Big|_0^m = m^3 - m^2 + m$.
Theo đề bài: $m^3 - m^2 + m = 6 \Leftrightarrow m^3 - m^2 + m - 6 = 0$.
Xét $f(m) = m^3 - m^2 + m - 6$. Ta thấy $f(2) = 8 - 4 + 2 - 6 = 0$. Vậy $m=2$ là một nghiệm.
Phân tích thành nhân tử: $m^3 - m^2 + m - 6 = (m-2)(m^2 + m + 3) = 0$.
Phương trình $m^2 + m + 3 = 0$ vô nghiệm vì $\Delta = 1 - 4*3 = -11 < 0$.
Vậy $m = 2$. Suy ra $m \in (0;4)$.

Ta có: $\displaystyle\int\limits_{a}^{a+1}{{{x}^{2}}\mathrm{d}x} = \dfrac{x^3}{3} |^{a+1}_{a} = \dfrac{(a+1)^3}{3} - \dfrac{a^3}{3} = \dfrac{{{\left(a+1 \right)}^{3}}-{{a}^{3}}}{3}$

Vậy đáp án đúng là $\dfrac{{{\left(a+1 \right)}^{3}}-{{a}^{3}}}{3}$

Ta có: $\dfrac{1}{\sqrt{x+1} - \sqrt{x}} = \dfrac{\sqrt{x+1} + \sqrt{x}}{(\sqrt{x+1} - \sqrt{x})(\sqrt{x+1} + \sqrt{x})} = \dfrac{\sqrt{x+1} + \sqrt{x}}{x+1 - x} = \sqrt{x+1} + \sqrt{x}$

Do đó: $I = \displaystyle\int\limits_{1}^{2} (\sqrt{x+1} + \sqrt{x}) dx = \displaystyle\int\limits_{1}^{2} (x+1)^{\frac{1}{2}} dx + \displaystyle\int\limits_{1}^{2} x^{\frac{1}{2}} dx$

$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3^{\frac{3}{2}} - 2^{\frac{3}{2}}) + \dfrac{2}{3}(2^{\frac{3}{2}} - 1^{\frac{3}{2}}) = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) + \dfrac{2}{3}(2\sqrt{2} - 1) = 2\sqrt{3} - \dfrac{2}{3} + \dfrac{4\sqrt{2}}{3} - \dfrac{4\sqrt{2}}{3} = 2\sqrt{3} + \dfrac{6}{3} - \dfrac{2}{3} = 2\sqrt{3} + \dfrac{4}{3}$

$I = \dfrac{2}{3} ( (x+1)\sqrt{x+1} )\Big|_1^2 + \dfrac{2}{3} ( x\sqrt{x} )\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) + \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3} = 2\sqrt{3} - \dfrac{2}{3}$

Tuy nhiên không có đáp án nào trùng khớp, ta kiểm tra lại đề bài, ta thấy có lẽ dấu trừ trong mẫu số phải là dấu cộng

$I = \displaystyle\int\limits_{1}^{2} \dfrac{1}{\sqrt{x+1}+\sqrt{x}}\mathrm{d}x = \displaystyle\int\limits_{1}^{2} (\sqrt{x+1} - \sqrt{x}) dx = \displaystyle\int\limits_{1}^{2} (x+1)^{\frac{1}{2}} dx - \displaystyle\int\limits_{1}^{2} x^{\frac{1}{2}} dx$

$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 - \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3^{\frac{3}{2}} - 2^{\frac{3}{2}}) - \dfrac{2}{3}(2^{\frac{3}{2}} - 1^{\frac{3}{2}}) = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) - \dfrac{2}{3}(2\sqrt{2} - 1) = 2\sqrt{3} - \dfrac{2}{3} - \dfrac{4\sqrt{2}}{3} + \dfrac{4\sqrt{2}}{3} = 2\sqrt{3} + \dfrac{6}{3} - \dfrac{2}{3} = 2\sqrt{3} - \dfrac{4}{3}$

$I = \dfrac{2}{3} ( (x+1)\sqrt{x+1} )\Big|_1^2 - \dfrac{2}{3} ( x\sqrt{x} )\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) - \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3} = 2\sqrt{3} + \dfrac{2}{3}$

Vậy đáp án đúng phải là $2\sqrt{3} + \dfrac{2}{3}$

Nếu đề là $I=\displaystyle\int\limits_{1}^{2} (\sqrt{x+1}+\sqrt{x}) dx$ thì ta làm như sau:

$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) + \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3} = 2\sqrt{3} - \dfrac{2}{3}$

Nếu đề là $I=\displaystyle\int\limits_{1}^{2} (\sqrt{x+1}-\sqrt{x}) dx$ thì ta làm như sau:

$I = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_1^2 - \dfrac{2}{3}x^{\frac{3}{2}}\Big|_1^2 = \dfrac{2}{3}(3\sqrt{3} - 2\sqrt{2}) - \dfrac{2}{3}(2\sqrt{2} - 1) = \dfrac{2}{3}3\sqrt{3} - \dfrac{2}{3}2\sqrt{2} - \dfrac{2}{3}2\sqrt{2} + \dfrac{2}{3} = 2\sqrt{3} + \dfrac{2}{3}$

Để ý nếu đổi cận từ 1->2 thành 0->1 thì:

$I=\displaystyle\int\limits_{0}^{1} (\sqrt{x+1}-\sqrt{x}) dx = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_0^1 - \dfrac{2}{3}x^{\frac{3}{2}}\Big|_0^1 = \dfrac{2}{3}(2\sqrt{2} - 1) - \dfrac{2}{3}(1 - 0) = \dfrac{4\sqrt{2}}{3} - \dfrac{4}{3}$

Nếu đổi cận từ 1->2 thành 0->1 thì:

$I=\displaystyle\int\limits_{0}^{1} (\sqrt{x+1}+\sqrt{x}) dx = \dfrac{2}{3}(x+1)^{\frac{3}{2}}\Big|_0^1 + \dfrac{2}{3}x^{\frac{3}{2}}\Big|_0^1 = \dfrac{2}{3}(2\sqrt{2} - 1) + \dfrac{2}{3}(1 - 0) = \dfrac{4\sqrt{2}}{3}$

Ta có: $I=\displaystyle\int\limits_{1}^{2}{\dfrac{1}{2}}\Big(\dfrac{1}{x}-\dfrac{1}{x+2} \Big)\mathrm{d}x = \dfrac{1}{2} \displaystyle\int\limits_{1}^{2} \Big(\dfrac{1}{x}-\dfrac{1}{x+2} \Big)\mathrm{d}x $
$\quad = \dfrac{1}{2} \Big[ \ln |x| - \ln |x+2| \Big]_{1}^{2} = \dfrac{1}{2} \Big[ \ln \Big|\dfrac{x}{x+2}\Big| \Big]_{1}^{2}$
$\quad = \dfrac{1}{2} \Big( \ln \dfrac{2}{4} - \ln \dfrac{1}{3} \Big) = \dfrac{1}{2} \Big( \ln \dfrac{1}{2} - \ln \dfrac{1}{3} \Big) $
$\quad = \dfrac{1}{2} (\ln 1 - \ln 2 - \ln 1 + \ln 3) = \dfrac{1}{2} (-\ln 2 + \ln 3) = -\dfrac{1}{2} \ln 2 + \dfrac{1}{2} \ln 3$
$\Rightarrow a = -\dfrac{1}{2}, b = \dfrac{1}{2}$
Vậy $T = a^2 + b^3 = \Big(-\dfrac{1}{2}\Big)^2 + \Big(\dfrac{1}{2}\Big)^3 = \dfrac{1}{4} + \dfrac{1}{8} = \dfrac{3}{8}$

Ta có: $\dfrac{x-2}{x+1} = 1 - \dfrac{3}{x+1}$.
Xét dấu: $x \in [1;2)$ thì $\dfrac{x-2}{x+1} < 0$; $x \in (2;5]$ thì $\dfrac{x-2}{x+1} > 0$.
Do đó:
$\displaystyle \int\limits_1^5 \left| \dfrac{x-2}{x+1} \right|\mathrm{d}x = -\int\limits_1^2 \dfrac{x-2}{x+1} dx + \int\limits_2^5 \dfrac{x-2}{x+1} dx$
$\displaystyle = -\int\limits_1^2 (1 - \dfrac{3}{x+1}) dx + \int\limits_2^5 (1 - \dfrac{3}{x+1}) dx$
$\displaystyle = -\left[ x - 3\ln|x+1| \right]_1^2 + \left[ x - 3\ln|x+1| \right]_2^5$
$\displaystyle = -\left[ (2 - 3\ln 3) - (1 - 3\ln 2) \right] + \left[ (5 - 3\ln 6) - (2 - 3\ln 3) \right]$
$\displaystyle = -1 + 3\ln 3 + 3\ln 2 + 3 - 3\ln 6$
$\displaystyle = 2 + 3\ln (\dfrac{6}{6}) = 2 + 3\ln 1 = 2$
Suy ra $a = 2, b = 0, c = 0$. Vậy $P = abc = 0$.