Q1: How does a romantic relationship affect student's academic perfomance

Q2: Is there a right age to fail in love? If yes when? If no why?

Q3: Should teenagers tell them parents alb their romantic relationship

Giups mình với càng dài càng tốt nha Thanksss

Cho hàm số\(f:R\rightarrow R\) thỏa mãn các tính chất sau:

\(\left\{{}\begin{matrix}f\left(x+y\right)=f\left(x\right)+f\left(y\right)+2xy\\f\left(\frac{1}{x}\right)=\frac{f\left(x\right)}{x^4}\end{matrix}\right.\)

Tính \(f\left(\sqrt{2019}\right)\)

Từ tập \(X=\left\{1;2;3;...;2020\right\}\) lấy ngẫu nhiên đồng thời hai số a và b. Tính xác suất sao cho hai số a và b lấy ra thỏa điều kiện " \(a^2+3b\) và \(b^2+3a\) đều là số chính phương

Ta có:

\(\left(1+1\right)^{2n+1}=C^0_{2n+1}+C^1_{2n+1}+...+C^{2n}_{2n+1}+C^{2n+1}_{2n}\)

\(=2\left(C^0_{2n+1}+C^1_{2n+1}+...+C^n_{2n+1}\right)\)

\(\Rightarrow\left(1+1\right)^{2n+1}=2.2^{20}\)

\(\Leftrightarrow n=10\)

\(sin3x+\sqrt{3}cos3x=2sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=\frac{1}{2}sin3x+\frac{\sqrt{3}}{2}cos3x\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(x+\frac{\pi}{3}\right)\)

\(\Leftrightarrow x\in R\)

Gọi H là trung điểm AB. Ta có:

\(E_A=E_B=\frac{kq}{AM^2}=\frac{kq}{\sqrt{a^2+h^2}}\)

\(E_M=\sqrt{E_A^2+E_B^2+2E_AE_B.cos\left(AMB\right)}\)

\(=\sqrt{2E_A^2+2E_A^2.\left(2cos^2\alpha-1\right)}\)

\(=\sqrt{4E_A^2.cos^2\alpha}=2E_A.cos\alpha\)

\(=\frac{2kq}{a\sqrt{a^2+h^2}}\)

Dễ thấy\(E_M\) đạt gtln khi \(h=0\Leftrightarrow E_M=\frac{2kq}{a^2}\)

\(\sqrt{3}sinx=cos\left(\frac{3\pi}{2}-2x\right)\)

\(\Leftrightarrow\sqrt{3}sinx=-cos\left(\frac{\pi}{2}-2x\right)\)

\(\Leftrightarrow\sqrt{3}sinx=-sin2x\)

\(\Leftrightarrow2sinx.cosx+\sqrt{3}sinx=0\)

\(\Leftrightarrow sinx\left(2cosx+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cosx=-\frac{\sqrt{3}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{5\pi}{6}+k2\pi\\x=-\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

Do\(x\in\left[\frac{-3\pi}{2};-\pi\right]\)

\(\Leftrightarrow x=-\pi;x=\frac{-7\pi}{6};x=\frac{-5\pi}{6}\)

\(f\left(x\right)=\frac{2-x}{x+1}+2=\frac{x+4}{x+1}< 0\)

\(\Leftrightarrow-4< x< -1\)

\(\left\{{}\begin{matrix}x^2-3x-4\le0\left(1\right)\\x^3-3x\left|x\right|-m^2+6m>0\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow-1\le x\le4\)

TH1 : \(-1\le x< 0\)

\(\left(2\right)\Leftrightarrow x^3+3x^2-m^2+6m>0\)

\(\Leftrightarrow m^2-6m\le x^3+3x^2=2\)

\(\Leftrightarrow m^2-6m-2\le0\)

\(\Leftrightarrow3-\sqrt{11}\le m\le3+\sqrt{11}\)

TH2 : \(0\le x< 4\)

\(\Leftrightarrow x^3-3x^2-m^2+6m\ge0\)

\(\Leftrightarrow m^2-6m\le x^3-3x^2=16\)

\(\Leftrightarrow m^2-6m-16\le0\)

\(-2\le m\le8\)

Vậy \(-2\le m\le8\)