Áp dụng BĐT Cô-si:

\(3\left(a^2+4\right)\ge3.4a=12a\)

\(b^4+b^4+b^4+81\ge4\sqrt[4]{81b^{12}}=12b^3\)

Cộng vế:

\(3\left(a^2+b^4\right)+93\ge12\left(a+b^3\right)=384\)

\(\Rightarrow a^2+b^4\ge85\)

\(\Rightarrow P\ge85-19=66\)

\(P_{min}=66\) khi \(\left(a;b\right)=\left(2;3\right)\)

Câu 5: 

\(\Leftrightarrow-x^2+7x-9+2x-9=0\)

\(\Leftrightarrow x^2-9x+18=0\)

=>x=3

=>Chọn A

Câu 18: B

Câu 19: A

18B

19A

20D

Câu 38: A

Câu 37: C

1. have done

2. has written/ has not finished

3. left/ have never met

4. have you had...?

5. did you do/ did you play

6. bought/ has not worn

7. has taught/ graduated

8. Have you heard...?/ has been/ Have you read...?

9. got/ was/ went

10. earned/ has already spent

will go

A

a

11. \(I=\int\limits^2_1x\sqrt{x^2+1}dx\)

Đặt \(\sqrt{x^2+1}=t\Leftrightarrow x^2=t^2-1\Rightarrow xdx=tdt\) ; \(\left\{{}\begin{matrix}x=1\Rightarrow t=\sqrt{2}\\x=2\Rightarrow t=\sqrt{5}\end{matrix}\right.\)

\(I=\int\limits^{\sqrt{5}}_{\sqrt{2}}t.tdt=\int\limits^{\sqrt{5}}_{\sqrt{2}}t^2dt=\dfrac{1}{3}t^3|^{\sqrt{5}}_{\sqrt{2}}=\dfrac{1}{3}\left(5\sqrt{5}-2\sqrt{2}\right)\)

12. Đặt \(\sqrt[3]{8-4x}=t\Rightarrow x=\dfrac{8-t^3}{4}\Rightarrow dx=-\dfrac{3}{4}t^2dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=2\\x=2\Rightarrow t=0\end{matrix}\right.\)

\(I=\int\limits^0_2t.\left(-\dfrac{3}{4}t^2dt\right)=\dfrac{3}{4}\int\limits^2_0t^3dt=\dfrac{3}{16}t^4|^2_0=3\)

13. Đặt \(\sqrt{3-2x}=t\Rightarrow x=\dfrac{3-t^2}{2}\Rightarrow dx=-tdt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=\sqrt{3}\\x=1\Rightarrow t=1\end{matrix}\right.\)

\(I=\int\limits^1_{\sqrt{3}}\dfrac{-tdt}{t}=\int\limits^{\sqrt{3}}_1dt=t|^{\sqrt{3}}_1=\sqrt{3}-1\)

a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp