Theo nguyên lý chia kẹo Euler, ta có \(C_{5-1}^{3-1}=6\) cách trao giải thưởng

ĐKXĐ: \(sin2x\ne1\Rightarrow x\ne\dfrac{\pi}{4}+k\pi\)

\(\dfrac{2cos2x}{1-sin2x}=0\Rightarrow cos2x=0\)

\(\Rightarrow2x=\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

Kết hợp ĐKXĐ: \(\Rightarrow x=\dfrac{3\pi}{4}+k\pi\)

4.

\(u_{n+1}=\dfrac{u_n}{3u_n+1}\Rightarrow\dfrac{1}{u_{n+1}}=\dfrac{3u_n+1}{u_n}=\dfrac{1}{u_n}+3\)

Đặt \(\dfrac{1}{u_n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{u_1}=2\\v_{n+1}=v_n+3\end{matrix}\right.\)

\(\Rightarrow v_n\) là cấp số cộng với công bội \(d=3\)

\(\Rightarrow v_n=v_1+\left(n-1\right)d=2+3\left(n-1\right)=3n-1\)

\(\Rightarrow u_n=\dfrac{1}{v_n}=\dfrac{1}{3n-1}\)

6.

\(u_{n+1}=4u_n-3u_{n-1}\Leftrightarrow u_{n+1}-u_n=3\left(u_n-u_{n-1}\right)\)

Đặt \(u_{n+1}-u_n=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_2-u_1=3\\v_n=3v_{n-1}\end{matrix}\right.\)

\(\Rightarrow v_n\) là cấp số nhân với công bội \(q=3\)

\(\Rightarrow v_n=3.3^{n-1}=3^n\)

\(\Rightarrow u_{n+1}-u_n=3^n\)

\(\Rightarrow u_{n+1}-\dfrac{1}{2}.3^{n+1}=u_n-\dfrac{1}{2}.3^n\)

Đặt \(u_n-\dfrac{1}{2}.3^n=x_n\Rightarrow\left\{{}\begin{matrix}x_1=u_1-\dfrac{1}{2}.3^1=-\dfrac{1}{2}\\x_{n+1}=x_n\end{matrix}\right.\)

\(\Rightarrow x_{n+1}=x_n=x_{n-1}=...=x_1=-\dfrac{1}{2}\)

\(\Rightarrow u_n-\dfrac{1}{2}.3^n=-\dfrac{1}{2}\Rightarrow u_n=\dfrac{3^n-1}{2}\)

TL:

\(789302687-100=789302587\)

\(5639174929-649102=5638525827\)

_HT_

trên đầu á

số 0(chắc thế) xin T.I.C.H

Khó quá, mới lớp 2 thôi mà

\(\lim\limits_{x→0}\dfrac{\sqrt{x+1}-\sqrt[3]{x+1}}{x}\\ =\lim\limits_{x→0}\dfrac{\sqrt{x+1}-1}{x}+\lim\limits_{x→0}\dfrac{1-\sqrt[3]{x+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x+1-1}{x\left(\sqrt{x+1}+1\right)}+\lim\limits_{x\rightarrow0}\dfrac{1-\left(x+1\right)}{x\left(1+\sqrt[3]{x+1}+\left(\sqrt[3]{x+1}\right)^2\right)}\\ =\lim\limits_{x\rightarrow0}\dfrac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\dfrac{-1}{1+\sqrt[3]{x+1}+\left(\sqrt[3]{x+1}\right)^2}\)

\(=\dfrac{1}{\sqrt{1}+1}+\dfrac{-1}{1+\sqrt[3]{1}+\left(\sqrt[3]{1}\right)^2}\\ =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)

lớp 6 mà hỏi toán lớp 11

k đi

hok tốt

@@@@@@@@@@@@@@@@@@

đó ai trl đc