\(f\left(x\right).f\left(1-x\right)=e^{x^2-x}\Rightarrow f\left(0\right).f\left(1\right)=1\)

\(\Rightarrow f\left(0\right)=1\)

Đặt \(\left\{{}\begin{matrix}u=2x^3-3x^2\\dv=\dfrac{f'\left(x\right)}{f\left(x\right)}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=6x^2-6x\\x=ln\left[f\left(x\right)\right]\end{matrix}\right.\)

\(\Rightarrow I=ln\left[f\left(x\right)\right].\left(2x^3-3x^2\right)|^1_0-\int\limits^1_0\left(6x^2-6x\right).ln\left[f\left(x\right)\right]dx\)

\(=0-\int\limits^1_0\left(6x^2-6x\right).ln\left[f\left(x\right)\right]dx\)

Đặt \(u=1-x\Rightarrow x=1-u\Rightarrow dx=-du\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow u=1\\x=1\Rightarrow u=0\end{matrix}\right.\)

\(x-x^2=x\left(1-x\right)=\left(1-u\right)u\)

\(\Rightarrow I=6\int\limits^0_1u\left(1-u\right).ln\left[f\left(1-u\right)\right].\left(-du\right)=-6\int\limits^1_0\left(u^2-u\right).ln\left[f\left(1-u\right)\right]du\)

\(=-6\int\limits^1_0\left(x^2-x\right)ln\left[f\left(1-x\right)\right]dx\)

\(=-6\int\limits^1_0\left(x^2-x\right)ln\left[\dfrac{e^{x^2-x}}{f\left(x\right)}\right]dx\)

\(=-6\int\limits^1_0\left(x^2-x\right).\left[lne^{x^2-x}-ln\left[f\left(x\right)\right]\right]dx\)

\(=-6\int\limits^1_0\left(x^2-x\right)^2dx+6\int\limits^1_0\left(x^2-x\right)ln\left[f\left(x\right)\right]dx\)

\(=-6\int\limits^1_0\left(x^2-x\right)^2dx-I\)

\(\Rightarrow2I=-6\int\limits^1_0\left(x^2-x\right)^2dx=-\dfrac{1}{5}\)

\(\Rightarrow I=-\dfrac{1}{10}\)

giải giúp mình 5 câu này ạ

26.

Do \(\dfrac{x^3+3x^2-x-3}{\left(x^2+2x+3\right)^2}=\dfrac{\left(x+1\right)\left(x^2+2x+3\right)-6\left(x+1\right)}{\left(x^2+2x+3\right)^2}\)

\(=\dfrac{x+1}{x^2+2x+3}-\dfrac{3\left(2x+2\right)}{\left(x^2+2x+3\right)^2}\)

\(\Rightarrow I=\int\limits^1_0\dfrac{-3d\left(x^2+2x+3\right)}{\left(x^2+2x+3\right)^2}+\dfrac{1}{2}\int\limits^1_0\dfrac{d\left(x^2+2x+3\right)}{x^2+2x+3}\)

\(=\dfrac{3}{x^2+2x+3}|^1_0+\dfrac{1}{2}ln\left(x^2+2x+3\right)|^1_0\)

\(=-\dfrac{1}{2}+\dfrac{1}{2}ln2=\dfrac{1}{2}\left(ln2-1\right)\)

\(\Rightarrow2a+b=1+2=3\)

27.

\(\int\dfrac{2tanx}{cos^2x}dx=\int2tanx.d\left(tanx\right)=tan^2x\)

\(\Rightarrow tan^2x|^a_0=1\Rightarrow tan^2a-tan^20=1\)

\(\Rightarrow tan^2a=1\Rightarrow a=\pm\dfrac{\pi}{4}\)

\(\int\dfrac{3x-2}{\left(x-2\right)^2}dx\)

Đặt \(x-2=u\Rightarrow x=u+2\Rightarrow dx=du\)

\(I=\int\dfrac{3\left(u+2\right)-2}{u^2}du=\int\dfrac{3u+4}{u^2}du=\int\left(\dfrac{3}{u}+\dfrac{4}{u^2}\right)du\)

\(=3ln\left|u\right|-\dfrac{4}{u}+C\)

\(=3ln\left|x-2\right|-\dfrac{4}{x-2}+C\)

Lời giải:
Gọi $z=a+bi$ với $a,b$ là số thực.

$|3-4i-z|=\sqrt{2}$

$\Leftrightarrow |(3-a)-i(4+b)|=\sqrt{2}$

$\Leftrightarrow (3-a)^2+(b+4)^2=2(*)$

$(z+3i)^2=[a+i(b+3)]^2=a^2-(b+3)^2+2ai(b+3)$ là số thuần ảo

Điều này xảy ra khi $a^2-(b+3)^2=0$

$\Leftrightarrow a^2=(b+3)^2\Rightarrow a=b+3$ hoặc $a=-(b+3)$
Nếu $a=b+3$ thì thay vào $(*)$ thì:

$b^2+(b+4)^2=2$

$\Leftrightarrow 2b^2+8b+14=0$

$\Leftrightarrow b^2+4b+7=0$

$\Leftrightarrow (b+2)^2=-3<0$ (vô lý - loại)

Nếu $a=-(b+3)$ thì thay vào $(*)$ thì:

$(b+6)^2+(b+4)^2=2$

$\Leftrightarrow 2b^2+20b+50=0$

$\Leftrightarrow b^2+10b+25=0$

$\Leftrightarrow (b+5)^2=0\Leftrightarrow b=-5$

Vậy có 1 số $b$ thỏa mãn kéo theo có 1 số $z$ thỏa mãn.

2259 là kết quả đúng. Nếu máy bấm ra 2250 thì là do máy tính sai.

Giả thiết tương đương:

\(x.f\left(x\right)=\sqrt{x^2+1}-\left(x^2+1\right)f'\left(x\right)\)

\(\Leftrightarrow\left(x^2+1\right)f'\left(x\right)+x.f\left(x\right)=\sqrt{x^2+1}\)

\(\Leftrightarrow\sqrt{x^2+1}.f'\left(x\right)+\dfrac{x}{\sqrt{x^2+1}}f\left(x\right)=1\)

\(\Leftrightarrow\left[\sqrt{x^2+1}.f\left(x\right)\right]'=1\)

Lấy nguyên hàm 2 vế:

\(\Rightarrow\sqrt{x^2+1}.f\left(x\right)=\int1.dx=x+C\)

Thay \(x=0\Rightarrow1.f\left(0\right)=0+C\Rightarrow C=1\)

\(\Rightarrow\sqrt{x^2+1}.f\left(x\right)=x+1\)

\(\Rightarrow f\left(x\right)=\dfrac{x+1}{\sqrt{x^2+1}}\)

\(\Rightarrow f\left(1\right)=\sqrt{2}\)

Chọn A