Đáp án:

\(\frac{(f'(x) (f(x))^2)^2}{1 + (f(x))^3} = \frac{1}{x+1}\)

\(\Rightarrow \frac{(3f'(x)f(x)^2)^2}{1 + (f(x))^3} = \frac{9}{x+1} \) (*)

Đặt \(h(x) = 1 + (f(x))^3 \Rightarrow  h'(x) = 3f'(x)(f(x))^2\)

(*) \(\Leftrightarrow \frac{(h'(x))^2}{h(x)} = \frac{9}{x+1} \Leftrightarrow  \frac{h'(x)}{\sqrt{h(x)}} =  \frac{3}{\sqrt{x+1}}\)

\(\Rightarrow \int \frac{h'(x)}{\sqrt{h(x)}} dx = \int \frac{3}{\sqrt{x+1}} dx \Rightarrow 2\sqrt{h(x)} = 6\sqrt{x+1} + C \quad \Rightarrow \quad h(x)= 9\sqrt{x+1} + C.\)

\(h(0) = 1 + (f(0))^3 = 9 \Rightarrow C = 0.\)

\(\Rightarrow h(x) = 9(x+1) \quad \Rightarrow \quad 1 + (f(x))^3 = 9x + 9.\)

\(\Rightarrow f(x) = \sqrt[3]{9x + 8}.\)

\(f(1) = \sqrt[3]{17} \approx 2.571 \quad \Rightarrow \boxed{\text{B}}.\)

page41

Đáp án:

\(f(x) = x f'(x) - 2x^3 - 3x^2 \quad \) Chia hai vế cho \(x^2\)

\(\Rightarrow \frac{f'(x)}{x} = \frac{f(x)}{x^2} + 2x + 3 \Leftrightarrow \frac{x f'(x) - f(x)}{x^2} = 2x + 3\)

\(\Leftrightarrow \left(\frac{f(x)}{x}\right)' = 2x + 3  \Leftrightarrow \frac{f(x)}{x} = x^2 + 3x + C\)

\(f(1) = 4 \Rightarrow C = 0\)

\(\Rightarrow  f(x) = x^3 + 3x^2 \Rightarrow  f(2) = 20 \quad \Rightarrow \boxed{\text{B}}\)

Gợi ý: \( \left(\frac{f(x)}{x}\right)' = \frac{x f'(x) - f(x)}{x^2}\)

page42

Đáp án:

Gợi ý: \(\left(\frac{f(x)}{e^x}\right)' = \frac{f'(x)e^x - e^x f(x)}{(e^x)^2} = \frac{f'(x) - f(x)}{e^x}\)

\(f'(x) = f(x) + x^2 e^x + 1 \quad \Rightarrow \quad \frac{f'(x) - f(x)}{e^x} = x^2 + \frac{1}{e^x}\)

\(\Rightarrow \frac{f(x)}{e^x} = \frac{x^3}{3} - \frac{1}{e^x} + C\)

\(f(0) = -1 \Rightarrow C = 0\)

\(\Rightarrow f(x) = \frac{x^3}{3}e^x - 1 \Rightarrow f(3)  = 9e^3 - 1 \quad \Rightarrow \boxed{D}.\)

page43

Đáp án:

Gợi ý: \((\ln{x} f(x))' = \frac{1}{x} f(x) + \ln{x} f'(x)\)

\(f(x) + x\ln{x} f'(x) = x^2 + 1\)

\(\Rightarrow \frac{1}{x}f(x) + \ln{x}f'(x) = x + \frac{1}{x}\)

\(\Rightarrow \ln{x}f(x) = \frac{x^2}{2} + \ln{x} + C\)

\(f(e) = \frac{e^2}{2} + 1 \Rightarrow C = 0\)

\(\Rightarrow f(x) = \frac{x^2}{2\ln{x}} + 1  \Rightarrow f(e^2) = \frac{e^4}{4} + 1 \quad \Rightarrow \boxed{\text{D}}\)

page44

Đáp án:

\(x(x+1)f'(x) + f(x) = x(x+1) \quad \text{Chia 2 vế cho } (x+1)^2\)

\(\Rightarrow \frac{x}{x+1}f'(x) + \frac{f(x)}{(x+1)^2} = \frac{x}{x+1}\) (Gợi ý: \( \left[\frac{x}{x+1}\right]' = (1 - \frac{1}{(x+1)})' =\frac{1}{(x+1)^2}\) )

Vì: \(\left[\frac{x}{x+1}f(x)\right]' = \frac{x}{x+1}f'(x) + \frac{f(x)}{(x+1)^2}\)

\(\Rightarrow \left[\frac{x}{x+1}f(x)\right]' = \frac{x}{x+1}.\)

\(\Rightarrow \frac{x}{x+1}f(x) = \int \frac{x}{x+1} dx = x - \ln{|x+1|} + C.\)

\(f(1) = -2\ln{2} \quad \Rightarrow -\ln{2}  = 1-\ln{2} +C\quad \Rightarrow C = -1.\)

\(\Rightarrow \frac{x}{x+1}f(x) =  - \ln{|x+1|} +x.\)

\(\Rightarrow \frac{2}{3} f(2) = -\ln{3} + 1 \quad \Rightarrow f(2) = \frac{3}{2} - \frac{3}{2}\ln{2}.\)

\(\Rightarrow a^2 + b^2 = \frac{9}{2} \quad \Rightarrow \boxed{\text{B}}.\)

Gợi ý: Vì \(\left[\frac{x}{x+1}f(x)\right]' = \frac{1}{(x+1)^2}f(x) + \frac{x}{x+1}f'(x).\)

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