\(I=\int\sqrt{20-x^2}dx-\int x^2dx\)

Xet \(I_1=\int\sqrt{20-x^2}dx\)

\(x=\sqrt{20}\sin t\left(-\dfrac{\pi}{2}\le t\le\dfrac{\pi}{2}\right)\Rightarrow dx=\sqrt{20}\cos tdt\)

\(\Rightarrow I_1=\int\sqrt{20\cos^2t}.\sqrt{20}\cos tdt=20\int\cos^2t.dt=10\int dt+10\int\cos2t.dt=10t+5\sin2t+C\)

\(\Rightarrow I=10arc\sin\left(\dfrac{x}{\sqrt{20}}\right)+5\sin\left[2.arc\sin\left(\dfrac{x}{\sqrt{20}}\right)\right]-\dfrac{1}{3}x^3+C\)

P/s: Bạn tự thay cận vô ạ

\(\left\{{}\begin{matrix}u_1=a;u_2=b\\u_{n+2}=\dfrac{1}{2}u_{n+1}+\dfrac{1}{2}u_n\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u_1=a,u_2=b\\u_{n+2}+\dfrac{1}{2}u_{n+1}=u_{n+1}+\dfrac{1}{2}u_n\end{matrix}\right.\)

\(v_{n+1}=u_{n+1}+\dfrac{1}{2}u_n\Rightarrow\left\{{}\begin{matrix}v_2=u_2+\dfrac{1}{2}u_1=b+\dfrac{1}{2}a\\v_{n+1}=v_n\end{matrix}\right.\)

\(\Rightarrow v_{n+1}=b+\dfrac{1}{2}a\Rightarrow u_{n+1}=b+\dfrac{1}{2}a-\dfrac{1}{2}u_n\)

\(\Leftrightarrow u_{n+1}-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)=-\dfrac{1}{2}\left[u_n-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)\right]\)

\(t_n=u_n-\left(\dfrac{1}{3}a+\dfrac{2}{3}b\right)\Rightarrow\left\{{}\begin{matrix}t_1=u_1-\dfrac{1}{3}a-\dfrac{2}{3}b=\dfrac{2}{3}\left(a-b\right)\\t_{n+1}=-\dfrac{1}{2}t_n\end{matrix}\right.\)

\(\Rightarrow t_n=\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}\Rightarrow u_n=t_n+\dfrac{1}{3}a+\dfrac{2}{3}b=\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\)

\(\Rightarrow limun=\lim\limits\left[\dfrac{2}{3}\left(a-b\right)\left(-\dfrac{1}{2}\right)^{n-1}+\dfrac{1}{3}a+\dfrac{2}{3}b\right]=0\)

\(S=5.\int\sqrt{\left(1-\dfrac{x}{8}\right)\left(1+\dfrac{x}{8}\right)}dx\)

\(t=1-\dfrac{x}{8}\Rightarrow x=8\left(1-t\right)\Rightarrow dx=-8dt\)

\(\Rightarrow S=-5.8\int\sqrt{t\left(1+\dfrac{8\left(1-t\right)}{8}\right)}dt=-40\int\sqrt{t\left(2-t\right)}dt=-40\int\sqrt{1-\left(t-1\right)^2}dt\)

\(t-1=\sin u\left(-\dfrac{\pi}{2}\le u\le\dfrac{\pi}{2}\right)\Rightarrow dt=\cos udu\)

\(\Rightarrow S=-40\int\cos^2u.du=-20\int[1+\cos\left(2u\right)]du\)

\(=-20\int du-20\int\cos\left(2u\right)du=-20u+\dfrac{20}{2}\sin2u=-20arc\sin\left(t-1\right)+10\sin2\left[arc\sin\left(t-1\right)\right]\)

\(=-20arc\sin\left(\dfrac{x}{8}\right)+10\sin2\left[arc\sin\left(\dfrac{x}{8}\right)\right]\)

P/s: Bạn tự thay cận vô ạ

\(a=-1< 0;\Delta=\left(2\sqrt{m}-1\right)^2+4\left(\sqrt{m}-m\right)=4m-4\sqrt{m}+1+4\sqrt{m}-4m=1>0\)

a/ \(f\left(x\right)\ge0\) vô nghiệm \(\Leftrightarrow f\left(x\right)< 0,\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a=-1< 0\left(tm\right)\\\Delta< 0\left(voly\right)\end{matrix}\right.\)

Vậy ko tồn tại m để ....

b/ \(f\left(x\right)\ge0,\forall x\in\left[1;2\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\\left[{}\begin{matrix}1< x_1< x_2\\x_1< x_2< 2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-1.f\left(1\right)>0\\\dfrac{x_1+x_2}{2}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}-1.f\left(2\right)>0\\\dfrac{x_1+x_2}{2}-2< 0\end{matrix}\right.\end{matrix}\right.\)

\(\left(1\right)\left\{{}\begin{matrix}-1+2\sqrt{m}-1-m+\sqrt{m}< 0\\\sqrt{m}-\dfrac{1}{2}-1>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m-3\sqrt{m}+2>0\\\sqrt{m}>\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}0< m< 1\\m>2\end{matrix}\right.\\m>\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow m>\dfrac{9}{4}\)

\(\left(2\right)\left\{{}\begin{matrix}-4+4\sqrt{m}-2-m+\sqrt{m}< 0\\\sqrt{m}-\dfrac{1}{2}-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m-5\sqrt{m}+6>0\\\sqrt{m}< \dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}0< m< 2\\m>3\end{matrix}\right.\\0\le m< \dfrac{25}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< m< 2\\3< m< \dfrac{25}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}m>\dfrac{9}{4}\\0< m< 2\\3< m< \dfrac{25}{4}\end{matrix}\right.\)

a/ \(\left(2-3i\right)\left(2+3i\right)=4-9i^2=4+9=13\)

b/ \(\left(2+3i\right)\left(4-i\right)=8-2i+12i-3i^2=11+10i\)

c/ \(=\dfrac{\left(1+i\right)\left(2+3i\right)}{4-9i^2}=\dfrac{2+5i-3}{4+9}=\dfrac{5i-1}{13}\)

\(I=\int\dfrac{\left(\sqrt{x+1}+\sqrt{x}\right)dx}{x+1-x}=\int\sqrt{x+1}dx+\int\sqrt{x}dx\)

Xet \(I_1=\int\sqrt{x+1}dx\)

\(t=x+1\Rightarrow dt=dx\Rightarrow I_1=\int\sqrt{t}.dt=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+C\)

\(\Rightarrow I=\dfrac{2}{3}\left(x+1\right)^{\dfrac{3}{2}}+\dfrac{2}{3}x^{\dfrac{3}{2}}+C\)

P/s: Bạn tự thay cận vô ạ

Bạn kiểm tra lại đề bài câu a, n=1 thì VT=1, VP=-1, nếu đề bài đúng thì vp phải là \(\dfrac{-\left(1\right)^{n-1}n\left(n+1\right)}{2}\)

\(n=1\Rightarrow VT=1=VP\)

Vậy mệnh đề đúng với n=1

Giả sử mệnh đề cũng đúng với \(n=k\left(k\in N\right)\), hay:

\(1^2-2^2+...+\left(-1\right)^{k-1}.k^2=\dfrac{\left(-1\right)^{k-1}.k.\left(k+1\right)}{2}\)

Ta cần chứng minh mệnh đề cũng đúng với \(n=k+1\) ,hay:

\(1^2-2^2+...+\left(-1\right)^{k-1}.k^2+\left(-1\right)^k.\left(k+1\right)^2=\dfrac{\left(-1\right)^k.\left(k+1\right)\left(k+2\right)}{2}\)

That vay:

\(VT=\dfrac{\left(-1\right)^{k-1}k\left(k+1\right)}{2}+\left(-1\right)^k.\left(k+1\right)^2=\dfrac{\left(-1\right)^{k-1}.k.\left(k+1\right)+2\left(-1\right)^k\left(k+1\right)^2}{2}\)

\(=\dfrac{\left(-1\right)^k\left(k+1\right)\left(-k+2k+2\right)}{2}=\dfrac{\left(-1\right)^k\left(k+1\right)\left(k+2\right)}{2}=VP\)

Vậy mệnh đề đúng với \(\forall n\in N\)

b/ \(n=7\Rightarrow VT=3^7=2187< 7!=5040\)

Vậy mệnh đề đúng với n=7

Giả sử mệnh đề đúng với \(n=k\left(k\in N;k\ge7\right)\),hay:

\(3^k\le k!\)

Ta cần chứng minh mệnh đề cũng đúng với \(n=k+1\) ,hay:

\(3^{k+1}\le\left(k+1\right)!\)

That vay:

\(3^k.3\le\left(k+1\right).k!\)

\(k>6\Rightarrow k+1>6\Rightarrow k+1>3\)

Ma \(3^k\le k!\)

\(\Rightarrow3^k.3\le\left(k+1\right).k!\Leftrightarrow3^{k+1}\le\left(k+1\right)!\)

Vậy mệnh đề đúng với \(\forall n\in N;n>6\)

Một cách làm khác đến từ vị trí của dân chuyên Toán :v Hãi hơn cái cách mình làm bao nhiêu ra. À bạn ấy làm từ cái tính nguyên hàm \(\int x.\dfrac{2x^4+1}{\sqrt{x^4+1}}dx\) trở đi nhá!