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Câu 1:

\(2f\left(x\right)+3f\left(\frac{2}{3x}\right)=5x\) (1)

Đặt \(t=\frac{2}{3x}\Rightarrow x=\frac{2}{3t}\)

\(\Rightarrow2f\left(\frac{2}{3t}\right)+3f\left(t\right)=5.\frac{2}{3t}\Leftrightarrow2f\left(\frac{2}{3t}\right)+3f\left(t\right)=\frac{10}{3t}\)

\(\Rightarrow2f\left(\frac{2}{3x}\right)+3f\left(x\right)=\frac{10}{3x}\Leftrightarrow3f\left(\frac{2}{3x}\right)+\frac{9}{2}f\left(x\right)=\frac{5}{x}\) (2)

Trừ vế cho vế của (2) cho (1):

\(\frac{5}{2}f\left(x\right)=\frac{5}{x}-5x\Rightarrow f\left(x\right)=\frac{2}{x}-2x\)

\(\Rightarrow\int\limits^1_{\frac{2}{3}}\frac{f\left(x\right)}{x}dx=\int\limits^1_{\frac{2}{3}}\left(\frac{2}{x^2}-2\right)dx=\left(-\frac{2}{x}-2x\right)|^1_{\frac{2}{3}}=\frac{1}{3}\)

Câu 2:

\(3f\left(x\right)-4f\left(2-x\right)=-x^2-12x+16\) (1)

Đặt \(2-x=t\Rightarrow x=2-t\)

\(\Rightarrow3f\left(2-t\right)-4f\left(t\right)=-\left(2-t\right)^2-12\left(2-t\right)+16\)

\(\Rightarrow3f\left(2-t\right)-4f\left(t\right)=-t^2+16t-12\)

\(\Rightarrow3f\left(2-x\right)-4f\left(x\right)=-x^2+16x-12\)

\(\Rightarrow4f\left(2-x\right)-\frac{16}{3}f\left(x\right)=-\frac{4}{3}x^2+\frac{64}{3}x-16\) (2)

Cộng (1) và (2):

\(-\frac{7}{3}f\left(x\right)=-\frac{14}{3}x^2+\frac{28}{3}x\)

\(\Rightarrow f\left(x\right)=2x^2-4x\)

\(\Rightarrow\int\limits^2_0f\left(x\right)dx=\int\limits^2_0\left(2x^2-4x\right)dx=-\frac{8}{3}\)

\(3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx\le2\int\limits^1_0\sqrt{f'\left(x\right)}f\left(x\right)dx\) (1)

Ta lại có:

\(3f'\left(x\right).f^2\left(x\right)+\frac{1}{3}\ge2\sqrt{f'\left(x\right)}.f\left(x\right)\)

\(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]\ge2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\) (2)

Từ (1); (2) \(\Rightarrow3\int\limits^1_0\left[f'\left(x\right).f^2\left(x\right)+\frac{1}{9}\right]dx=2\int\limits^1_0\sqrt{f'\left(x\right)}.f\left(x\right)dx\)

Dấu "=" xảy ra khi và chỉ khi:

\(3f'\left(x\right).f^2\left(x\right)=\frac{1}{3}\Rightarrow3\int f'\left(x\right).f^2\left(x\right)dx=\int\frac{1}{3}dx\)

\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+C\)

Thay \(x=0\Rightarrow f^3\left(0\right)=C\Rightarrow C=1\)

\(\Rightarrow f^3\left(x\right)=\frac{x}{3}+1\Rightarrow\int\limits^1_0f^3\left(x\right)dx=\int\limits^1_0\left(\frac{x}{3}+1\right)dx=\frac{7}{6}\)

Câu 1:

\(\int\limits^3_0\left(f'\left(x\right)+1\right)\sqrt{x+1}dx=\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx+\int\limits^3_0\sqrt{x+1}dx\)

\(=\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx+\frac{14}{3}=\frac{302}{15}\Rightarrow\int\limits^1_0f'\left(x\right)\sqrt{x+1}dx=\frac{232}{15}\)

Ta có:

\(I=\int\limits^3_0\frac{f\left(x\right)dx}{\sqrt{x+1}}\)

Đặt \(\left\{{}\begin{matrix}u=f\left(x\right)\\dv=\frac{dx}{\sqrt{x+1}}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=f'\left(x\right)dx\\v=2\sqrt{x+1}\end{matrix}\right.\)

\(\Rightarrow I=2f\left(x\right)\sqrt{x+1}|^3_0-2\int\limits^3_0f'\left(x\right)\sqrt{x+1}dx\)

\(=4f\left(3\right)-2f\left(0\right)-2.\frac{232}{15}\)

\(=2\left(2f\left(3\right)-f\left(0\right)\right)-\frac{464}{15}=36-\frac{464}{15}=\frac{76}{15}\)

Câu 2:

\(I_1=\int\limits^3_1\frac{xf'\left(x\right)}{x+1}dx=0\)

Đặt \(\left\{{}\begin{matrix}u=\frac{x}{x+1}\\dv=f'\left(x\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{1}{\left(x+1\right)^2}dx\\v=f\left(x\right)\end{matrix}\right.\)

\(\Rightarrow I_1=\frac{xf\left(x\right)}{x+1}|^3_1-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}=\frac{3.3}{3+1}-\frac{1.3}{1+1}-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=\frac{3}{4}-\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=0\)

\(\Rightarrow\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx=\frac{3}{4}\)

Ta có:

\(I=\int\limits^3_1\frac{f\left(x\right)+lnx}{\left(x+1\right)^2}dx=\int\limits^3_1\frac{f\left(x\right)}{\left(x+1\right)^2}dx+\int\limits^3_1\frac{lnx}{\left(x+1\right)^2}dx=\frac{3}{4}+I_2\)

Xét \(I_2=\int\limits^3_1\frac{lnx}{\left(x+1\right)^2}dx\Rightarrow\) đặt \(\left\{{}\begin{matrix}u=lnx\\dv=\frac{1}{\left(x+1\right)^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x}\\v=\frac{-1}{x+1}\end{matrix}\right.\)

\(\Rightarrow I_2=\frac{-lnx}{x+1}|^3_1+\int\limits^3_1\frac{dx}{x\left(x+1\right)}=-\frac{1}{4}ln3+\int\limits^1_0\left(\frac{1}{x}-\frac{1}{x+1}\right)dx\)

\(=-\frac{1}{4}ln3+ln\left(\frac{x}{x+1}\right)|^3_1=-\frac{1}{4}ln3+ln\frac{3}{4}-ln\frac{1}{2}=\frac{3}{4}ln3-ln2\)

\(\Rightarrow I=\frac{3}{4}+\frac{3}{4}ln3-ln2\)

\(\int\limits^2_0x.f'\left(2x\right)dx\) thì tính được, còn \(\int\limits^4_0x.f'\left(2x\right)dx\) thì mình nghĩ thế này chưa đủ dữ liệu để tính