Câu 2:

a, Số hạt p, n, e lần lượt là:

- Trong nitrogen: 7,7,7

- Trong fluorine: 9, 10, 9

- Trong neon: 10, 10, 10

b, Hai hạt luôn có số lượng bằng nhau là: p và e

a.

Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{t^2-1}{2}\)

Phương trình trở thành:

\(2t+t^2-1+1=0\)

\(\Rightarrow t\left(t+2\right)=0\Rightarrow\left[{}\begin{matrix}t=0\\t=-2< -\sqrt{2}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx+cosx=0\)

\(\Rightarrow tanx=-1\)

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

a, Đặt \(sinx+cosx=t\left(t\in\left[-\sqrt{2};\sqrt{2}\right]\right)\)

\(pt\Leftrightarrow2t+t^2-1+1=0\)

\(\Leftrightarrow t^2+2t=0\)

\(\Leftrightarrow t\left(t+2\right)=0\)

\(\Leftrightarrow t=0\)

\(\Leftrightarrow sinx+cosx=0\)

\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\)

b.

Đặt \(sinx-cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow sinx.cosx=\dfrac{1-t^2}{2}\)

Phương trình trở thành:

\(2\sqrt{2}t-2\left(1-t^2\right)=1\)

\(\Leftrightarrow2t^2+2\sqrt{2}t-3=0\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{2}}{2}\\t=-\dfrac{3\sqrt{2}}{2}< -\sqrt{2}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{5}+k2\pi\end{matrix}\right.\)

\(\Rightarrow...\)

#include <bits/stdc++.h>

using namespace std;

long long n,i,s,x;

int main()

{

cin>>n;

s=1;

for (i=1; i<=n; i++)

{

cin>>x;

if (x%2!=0) s=s*x;

}

cout<<s;

return 0;

}

1.

\(cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{\pi}{3}+x\right)+1=0\)

\(\Leftrightarrow2cos^2\left(\dfrac{\pi}{3}+x\right)+cos\left(\dfrac{\pi}{3}+x\right)=0\)

\(\Leftrightarrow cos\left(\dfrac{\pi}{3}+x\right)\left[2cos\left(\dfrac{\pi}{3}+x\right)+1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(\dfrac{\pi}{3}+x\right)=0\\cos\left(\dfrac{\pi}{3}+x\right)=-\dfrac{1}{2}\end{matrix}\right.\)

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

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

ok

hay quá bn ơiiiii

Hay ta!

\(\lim\limits_{x\rightarrow5}\left(x^3+5x^2-10x+8\right)=5^3+5.5^2-10.5+8=...\)

\(\lim\limits_{x\rightarrow-2}\dfrac{x^3-x^2-2x-8}{x^2+3x+2}=\dfrac{-16}{0}=-\infty\)

\(\lim\limits_{x\rightarrow-\infty}\dfrac{x^2-5x+2}{2\left|x\right|+1}=\lim\dfrac{\left|x\right|-5+\dfrac{2}{\left|x\right|}}{2+\dfrac{1}{\left|x\right|}}=\dfrac{+\infty}{2}=+\infty\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt[3]{x^3+4x-3}-4x}{\sqrt{9x^2-5x+1}-4x}=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(\sqrt[3]{1+\dfrac{4}{x^2}-\dfrac{3}{x^3}}-4\right)}{x\left(\sqrt[]{9-\dfrac{5}{x}+\dfrac{1}{x^2}}-4\right)}=\dfrac{1-4}{3-4}=3\)

Lời giải:

a.

\(\lim\limits_{x\to 5}(x^3+5x^2-10x+8)=5^3+5.5^2-10.5+8=208\)

b. 

\(L=\lim\limits_{x\to -2}\frac{x^3-x^2-2x-8}{x^2+3x+2}\lim\limits_{x\to -2}\frac{x^3-x^2-2x-8}{x+1}.\frac{1}{x+2}=16\lim\limits_{x\to -2}\frac{1}{x+2}\)\(\lim\limits_{x\to -2-}\frac{1}{x+2}=-\infty \Rightarrow L=-\infty ; \lim\limits_{x\to -2+}\frac{1}{x+2}=+\infty \Rightarrow L=+\infty \)

c.

\(\lim\limits_{x\to -\infty}\frac{x^2-5x+2}{2|x|+1}=\lim\limits_{x\to -\infty}\frac{|x|-\frac{5x}{|x|}+\frac{2}{|x|}}{2+\frac{1}{|x|}}\)

\(=\lim\limits_{x\to -\infty}\frac{|x|-\frac{5x}{-x}}{2}=\frac{1}{2}\lim\limits_{x\to -\infty}(|x|+5)=+\infty \)

a) \(\left(2m-1\right)sinx+1-m=0\Rightarrow sinx=\dfrac{m-1}{2m-1}\)

     Pt có nghiệm:  \(-1\le\dfrac{m-1}{2m-1}\le1\)

                           \(\Rightarrow1-2m\le m-1\le2m-1\Rightarrow m\ge\dfrac{2}{3}\)

b) \(\left(m+1\right)sin3x-cos3x=m+2\)

    Pt có nghiệm:   \(\left(m+1\right)^2+\left(-1\right)^2\ge\left(m+2\right)^2\)

                           \(\Rightarrow m^2+2m+1+1\ge m^2+4m+4\)

                           \(\Rightarrow-2m\ge2\Rightarrow m\le-1\)

a, \(\left(2m-1\right)sinx+1-m=0\)

\(\Leftrightarrow sinx=\dfrac{m-1}{2m-1}\)

Phương trình có nghiệm khi:

\(-1\le\dfrac{m-1}{2m-1}\le1\)

\(\Leftrightarrow\left[{}\begin{matrix}m>\dfrac{2}{3}\\m\le0\end{matrix}\right.\)