Câu 9: C

Câu 10: A

9 c

10 a

c

a

a: AC=4cm

AH=2,4cm

BH=1,8cm

CH=3,2cm

Chọn B

B

B

9.B

10.A

11.D

\(9,\tan B=\dfrac{AC}{AB}=\dfrac{4}{3}\left(B\right)\\ 10,A=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\dfrac{\sqrt{x}+4}{x+16}\\ A=\dfrac{x+16}{\left(\sqrt{x}-4\right)\left(x+16\right)}=\dfrac{1}{\sqrt{x}-4}\left(A\right)\\ 11,B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(x-1\right)^2}{2}\\ B=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\\ B=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\\ B>0\Leftrightarrow\sqrt{x}-1< 0\left(-\sqrt{x}< 0\right)\\ \Leftrightarrow0\le x< 1\left(D\right)\)

a

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Bài 3.9:

a)

\(\int ^{1}_{0}(y^3+3y^2-2)dy=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{y^4}{4}+y^3-2y \right )=\frac{-3}{4}\)

b) \(\int ^{4}_{1}\left (t+\frac{1}{\sqrt{t}}-\frac{1}{t^2}\right)dt=\left.\begin{matrix} 4\\ 1\end{matrix}\right|\left ( \frac{t^2}{2}+2\sqrt{t}+\frac{1}{t} \right )=\frac{35}{4}\)

d) Ta có:

\(\int ^{1}_{0}(3^s-2^s)^2ds=\int ^{1}_{0}(9^s+4^s-2.6^s)ds=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\left ( \frac{9^s}{\ln 9}+\frac{4^s}{\ln 4}-\frac{2.6^s}{\ln 6} \right )\)

\(=\frac{8}{\ln 9}+\frac{3}{\ln 4}-\frac{10}{\ln 6}\)

h)

Ta có \(\int ^{\frac{5\pi}{4}}_{\pi}\frac{\sin x-\cos x}{\sqrt{1+\sin 2x}}dx=\int ^{\frac{5\pi}{4}}_{\pi}\frac{\sin x-\cos x}{\sqrt{\sin^2x+\cos^2x+2\sin x\cos x}}dx\)

\(=\int ^{\frac{5\pi}{4}}_{\pi}\frac{-d(\sin x+\cos x)}{|\sin x+\cos x|}=\int ^{\frac{5\pi}{4}}_{\pi}\frac{d(\sin x+\cos x)}{\sin x+\cos x}=\left.\begin{matrix} \frac{5\pi}{4}\\ \pi\end{matrix}\right|\ln |\sin x+\cos x|=\ln (\sqrt{2})\)

Bài 3.10:

a)

Đặt \(t=1-x\) thì:

\(\int ^{2}_{1}x(1-x)^5dx=\int ^{-1}_{0}t^5(1-t)d(1-t)=\int ^{0}_{-1}t^5(1-t)dt\)

\(=\left.\begin{matrix} 0\\ -1\end{matrix}\right|\left ( \frac{t^6}{6}-\frac{t^7}{7} \right )=\frac{-13}{42}\)

b) Đặt \(\sqrt{e^x-1}=t\) \(\Rightarrow x=\ln (t^2+1)\)

Khi đó

\(\int ^{\ln 2}_{0}\sqrt{e^x-1}dx=\int ^{1}_{0}td(\ln (t^2+1))=\int ^{1}_{0}t.\frac{2t}{t^2+1}dt\)

\(=\int ^{1}_{0}\frac{2t^2}{t^2+1}dt=\int ^{1}_{0}2dt-\int ^{1}_{0}\frac{2}{t^2+1}dt=\left.\begin{matrix} 1\\ 0\end{matrix}\right|2t-\int ^{1}_{0}\frac{2dt}{t^2+1}=2-\int ^{1}_{0}\frac{2dt}{t^2+1}\)

Với \(\int ^{1}_{0}\frac{2dt}{t^2+1}\), đặt \(t=\tan m\)

\(\Rightarrow \int ^{1}_{0}\frac{2dt}{t^2+1}=\int ^{\frac{\pi}{4}}_{0}\frac{2d(\tan m)}{\tan ^2m+1}=\int ^{\frac{\pi}{4}}_{0}2\cos ^2md(\tan m)\)

\(=\int ^{\frac{\pi}{4}}_{0}2dm=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|2m=\frac{\pi}{2}\)

Do đó \(\int ^{\ln 2}_{0}\sqrt{e^x-1}dx=2-\frac{\pi}{2}\)

Bài 3.10

c) Đặt \(t=\sqrt[3]{1-x}\Rightarrow x=1-t^3\)

\(\int ^{9}_{1}x\sqrt[3]{1-x}dx=\int ^{-2}_{0}(1-t^3)td(1-t^3)=\int ^{0}_{-2}3t^2.t(1-t^3)dt\)

\(\left.\begin{matrix} 0\\ -2\end{matrix}\right|\left ( \frac{3t^4}{4}-\frac{3t^7}{7} \right )=\frac{-468}{7}\)

e) Đặt \(t=\frac{1}{x}\) suy ra:

\(\int ^{2}_{1}\frac{\sqrt{x^2+1}}{x^4}dx=\int ^{\frac{1}{2}}_{1}t^4\sqrt{\frac{1}{t^2}+1}d\left(\frac{1}{t}\right)\)

\(=\int ^{1}_{\frac{1}{2}}\frac{t^4\sqrt{t^2+1}}{t}.\frac{1}{t^2}dt=\int ^{1}_{\frac{1}{2}}t\sqrt{t^2+1}dt=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\sqrt{t^2+1}d(t^2+1)\)

\(=\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left(\frac{1}{2}.\frac{2}{3}\sqrt{(t^2+1)^3}\right)=\frac{2\sqrt{2}}{3}-\frac{5\sqrt{5}}{24}\)