Bạn cần bài nào trong mấy bài này nhỉ?

1.

\(u_{n+1}=4u_n+3.4^n\)

\(\Leftrightarrow u_{n+1}-\dfrac{3}{4}\left(n+1\right).4^{n+1}=4\left[u_n-\dfrac{3}{4}n.4^n\right]\)

Đặt \(u_n-\dfrac{3}{4}n.4^n=v_n\Rightarrow\left\{{}\begin{matrix}v_1=2-\dfrac{3}{4}.4=-1\\v_{n+1}=4v_n\end{matrix}\right.\)

\(\Rightarrow v_n=-1.4^{n-1}\)

\(\Rightarrow u_n=\dfrac{3}{4}n.4^n-4^{n-1}=\left(3n-1\right)4^{n-1}\)

2.

\(a_n=\dfrac{a_{n-1}}{2n.a_{n-1}+1}\Rightarrow\dfrac{1}{a_n}=2n+\dfrac{1}{a_{n-1}}\)

\(\Leftrightarrow\dfrac{1}{a_n}-n^2-n=\dfrac{1}{a_{n-1}}-\left(n-1\right)^2-\left(n-1\right)\)

Đặt \(\dfrac{1}{a_n}-n^2-n=b_n\Rightarrow\left\{{}\begin{matrix}b_1=2-1-1=0\\b_n=b_{n-1}=...=b_1=0\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a_n}=n^2+n\Rightarrow a_n=\dfrac{1}{n^2+n}\)

a.

\(sin\left(2x-\dfrac{\pi}{4}\right)=-1\)

\(\Leftrightarrow2x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\) (1)

\(-\dfrac{\pi}{3}\le x\le\dfrac{7\pi}{3}\Rightarrow-\dfrac{\pi}{3}\le-\dfrac{\pi}{8}+k\pi\le\dfrac{7\pi}{3}\)

\(\Rightarrow-\dfrac{5}{24}\le k\le\dfrac{59}{24}\Rightarrow k=\left\{0;1;2\right\}\)

Thế vào (1) \(\Rightarrow x=\left\{-\dfrac{\pi}{8};\dfrac{7\pi}{8};\dfrac{15\pi}{8}\right\}\)

Câu b lm ntn ạ 

c.

\(\Leftrightarrow sin4x=sin\left(3x-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=3x-\dfrac{\pi}{2}+k2\pi\\4x=\dfrac{3\pi}{2}-3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{2}+k2\pi\\x=\dfrac{3\pi}{14}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

d.

\(\Leftrightarrow sin\left(2x+30^0\right)=sin\left(30^0+x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+30^0=30^0+x+k360^0\\2x+30^0=150^0-x+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k360^0\\x=40^0+k120^0\end{matrix}\right.\)

e.

\(\Leftrightarrow cos3x=-sinx\)

\(\Leftrightarrow cos3x=cos\left(\dfrac{\pi}{2}+x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{\pi}{2}+x+k2\pi\\3x=-\dfrac{\pi}{2}-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)

f.

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)\left(sin2x+cos5x\right)=0\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)\left(sin2x-sin\left(5x-\dfrac{\pi}{2}\right)\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(2x-\dfrac{\pi}{4}\right)=0\\sin\left(5x-\dfrac{\pi}{2}\right)=sin2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=k\pi\\5x-\dfrac{\pi}{2}=2x+k2\pi\\5x-\dfrac{\pi}{2}=\pi-2x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{3\pi}{14}+\dfrac{k2\pi}{7}\end{matrix}\right.\)

Tức là câu 2, 3 của bài hình không gian đúng không em?

Đúng rồi ạ , Thầy giúp em với ạ !

Gọi H là trung điểm AB, có lẽ từ 2 câu trên ta đã phải chứng minh được \(SH\perp\left(ABCD\right)\)

Do \(\left\{{}\begin{matrix}DM\cap\left(SAC\right)=S\\MS=\dfrac{1}{2}DS\end{matrix}\right.\) \(\Rightarrow d\left(M;\left(SAC\right)\right)=\dfrac{1}{2}d\left(D;\left(SAC\right)\right)\)

Gọi E là giao điểm AC và DH

Talet: \(\dfrac{HE}{DE}=\dfrac{AH}{DC}=\dfrac{1}{2}\Rightarrow HE=\dfrac{1}{2}DE\)

\(\left\{{}\begin{matrix}DH\cap\left(SAC\right)=E\\HE=\dfrac{1}{2}DE\end{matrix}\right.\) \(\Rightarrow D\left(H;\left(SAC\right)\right)=\dfrac{1}{2}d\left(D;\left(SAC\right)\right)=d\left(M;\left(SAC\right)\right)\)

Từ H kẻ HF vuông góc AC (F thuộc AC), từ H kẻ \(HK\perp SF\)

\(\Rightarrow HK\perp\left(SAC\right)\Rightarrow HK=d\left(H;\left(SAC\right)\right)\)

ABCD là hình vuông \(\Rightarrow\widehat{HAF}=45^0\Rightarrow HF=AH.sin45^0=\dfrac{a\sqrt{2}}{4}\)

\(SH=\dfrac{a\sqrt{3}}{2}\), hệ thức lượng:

\(HK=\dfrac{SH.HF}{\sqrt{SH^2+HF^2}}=\dfrac{a\sqrt{21}}{14}\)

\(\Rightarrow d\left(M;\left(SAC\right)\right)=\dfrac{a\sqrt{21}}{14}\)

j mà gửi lên lớp 11 vậy má!!!

ghi lun cái đề đi