Um

Mới thcs mà ế 11 năm (khê lắm)

a/ \(-1\le sin2x\le1\Rightarrow-7\le y\le-1\)

b/ \(-1\le cos2x\le1\Rightarrow1\le y\le7\)

c/ \(-1\le sin2x\le1\Rightarrow3\le y\le11\)

d/ \(-1\le cos\left(3x+\frac{\pi}{3}\right)\le1\Rightarrow8\le y\le12\)

= gì vậy ạ

Nếu

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+x+1}-2\sqrt{x^2-x+1}\right)=\lim\limits_{x\rightarrow+\infty}x\left(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}-2\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}\right)\)

\(=+\infty.\left(1-2\right)=-\infty\)

Nếu:

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+x+1}-2\sqrt{x^2-x+1}\right)=\lim\limits_{x\rightarrow-\infty}x\left(-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+2\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}\right)\)

\(=-\infty.\left(-1+2\right)=-\infty\)

\(\Leftrightarrow2\sqrt{3}sinx.cosx+2cos^2x-1=2cosx-1\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx+cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\\sqrt{3}sinx+cosx=1\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=\frac{1}{2}\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}=sin\left(\frac{\pi}{6}\right)\)

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

Phương trình lượng giác bậc nhất cơ bản mà :(

\(\Leftrightarrow\sin x-\cos x=\frac{\sqrt{2}}{2}\Leftrightarrow\sin\left(x-\frac{\pi}{4}\right)=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{6}+k2\pi\\x-\frac{\pi}{4}=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5\pi}{12}+k2\pi\\x=\frac{13}{12}\pi+k2\pi\end{matrix}\right.\)

\(th1:0\le\frac{5\pi}{12}+k2\pi\le2\pi\)

\(th2:0\le\frac{13}{12}\pi+k2\pi\le2\pi\)

Chặn k là okie :)

\(B=\lim\limits_{x\rightarrow+\infty}x\left(\frac{\left(\sqrt{x^2+2x}+x\right)^2-4\left(x^2+x\right)}{\sqrt{x^2+2x}+x+2\sqrt{x^2+x}}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}2x^2\left(\frac{\sqrt{x^2+2x}-x-1}{\sqrt{x^2+2x}+x+2\sqrt{x^2+x}}\right)\)

\(=\lim\limits_{x\rightarrow+\infty}\frac{2x^2\left(x^2+2x-\left(x+1\right)^2\right)}{\left(\sqrt{x^2+2x}+x+2\sqrt{x^2+x}\right)\left(\sqrt{x^2+2x}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\frac{-2x^2}{\left(\sqrt{x^2+2x}+x+2\sqrt{x^2+x}\right)\left(\sqrt{x^2+2x}+x+1\right)}\)

\(=\lim\limits_{x\rightarrow+\infty}\frac{-2x^2}{x^2\left(\sqrt{1+\frac{2}{x}}+1+2\sqrt{1+\frac{1}{x}}\right)\left(\sqrt{1+\frac{2}{x}}+1+\frac{1}{x}\right)}=\frac{-2}{\left(1+1+2\right)\left(1+1+0\right)}=-\frac{1}{4}\)

Hàm \(cos2x\) tuần hoàn chu kì \(\pi\) nên ta chỉ cần tìm số nghiệm của pt \(cos2x=\frac{\pi}{4}\) trên \(\left[0;\pi\right]\)

\(cos2x=\frac{\pi}{4}\Rightarrow x=\pm\frac{1}{2}arccos\left(\frac{\pi}{4}\right)+k\pi\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}arcos\left(\frac{\pi}{4}\right)\\x=\pi-\frac{1}{2}arccos\left(\frac{\pi}{4}\right)\end{matrix}\right.\)

Có 2 nghiệm trên \(\left[0;\pi\right]\Rightarrow\) có \(\left(2017+2017+1\right).2=8070\) nghiệm trên đoạn đã cho

bạn có thể giải thích chỗ dấu suy ra đc ko :((

2/

\(\Leftrightarrow3sinx-4sin^3x-\sqrt{3}cosx=2sinx\)

\(\Leftrightarrow4sin^3x-sinx+\sqrt{3}cosx=0\)

Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow4tan^3x-tanx\left(1+tan^2x\right)+\sqrt{3}\left(1+tan^2x\right)=0\)

\(\Leftrightarrow3tan^3x+\sqrt{3}tan^2x-tanx+\sqrt{3}=0\)

Bạn xem lại đề, pt bậc 3 này ko giải được (nghiệm rất xấu)

1.

\(\Leftrightarrow\sqrt{3}cos^2x-\sqrt{3}+cos^2x+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)

\(\Leftrightarrow-\sqrt{3}sin^2x+cosx+\left(\sqrt{3}-1\right)sinx.cosx+sinx-cosx=0\)

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+\sqrt{3}sinx-1\right)=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(\frac{1}{2}cosx+\frac{\sqrt{3}}{2}sinx-\frac{1}{2}\right)=0\)

\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\left[sin\left(x+\frac{\pi}{6}\right)-\frac{1}{2}\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{6}\right)=\frac{1}{2}\end{matrix}\right.\)