Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(y=sin2x+\left(\sqrt{3}+1\right)cos2x+sin^2x-cos^2x-1\)
\(=sin2x+\sqrt{3}cos2x-1=2sin\left(2x+\frac{\pi}{3}\right)-1\)
Do \(-1\le sin\left(2x+\frac{\pi}{3}\right)\le1\Rightarrow-3\le y\le1\)
b/ \(y=2sin^2x-2cos^2x-3sinx.cosx-1\)
\(=-2cos2x-\frac{3}{2}sin2x-1=-\frac{5}{2}\left(\frac{3}{5}sinx+\frac{4}{5}cosx\right)-1\)
\(=-\frac{5}{2}sin\left(x+a\right)-1\Rightarrow-\frac{7}{2}\le y\le\frac{3}{2}\)
c/ \(y=1-sin2x+2cos2x+\frac{3}{2}sin2x=\frac{1}{2}sin2x+2cos2x+1\)
\(=\frac{\sqrt{17}}{2}\left(\frac{1}{\sqrt{17}}sin2x+\frac{4}{\sqrt{17}}cos2x\right)+1=\frac{\sqrt{17}}{2}sin\left(2x+a\right)+1\)
\(\Rightarrow-\frac{\sqrt{17}}{2}+1\le y\le\frac{\sqrt{17}}{2}+1\)
3.
Theo điều kiện của pt lượng giác bậc nhất:
\(m^2+\left(3m+1\right)^2\ge\left(1-2m\right)^2\)
\(\Leftrightarrow10m^2+6m+1\ge4m^2-4m+1\)
\(\Leftrightarrow3m^2+5m\ge0\Rightarrow\left[{}\begin{matrix}m\ge0\\m\le-\frac{5}{3}\end{matrix}\right.\)
4.
\(\Leftrightarrow1-sin^2x-\left(m^2-3\right)sinx+2m^2-3=0\)
\(\Leftrightarrow-sin^2x-m^2sinx+2m^2+3sinx-2=0\)
\(\Leftrightarrow\left(-sin^2x+3sinx-2\right)+m^2\left(2-sinx\right)=0\)
\(\Leftrightarrow\left(sinx-1\right)\left(2-sinx\right)+m^2\left(2-sinx\right)=0\)
\(\Leftrightarrow\left(2-sinx\right)\left(sinx-1+m^2\right)=0\)
\(\Leftrightarrow sinx=1-m^2\)
\(\Rightarrow-1\le1-m^2\le1\)
\(\Rightarrow m^2\le2\Rightarrow-\sqrt{2}\le m\le\sqrt{2}\)
1.
Bạn xem lại đề, \(sin^2x\left(\frac{x}{2}-\frac{\pi}{4}\right)\) là sao nhỉ?Có cả x trong lẫn ngoài ngoặc?
2.
ĐKXĐ: \(sinx\ne0\)
\(\left(2sinx-cosx\right)\left(1+cosx\right)=sin^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)=1-cos^2x\)
\(\Leftrightarrow\left(2sinx-cosx\right)\left(1+cosx\right)-\left(1+cosx\right)\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1+cosx\right)\left(2sinx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=-1\\sinx=\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)
Khi cho A td KOH thu được ancol đồng đẳng. => Các ancol là no đơn chức mạch hở.
Gọi CT các este: \(C_mH_{2m+1}COOC_{m'}H_{2m'+1};C_nH_{2n-1}COOC_{n'}H_{2n'-1};C_qH_{2q}\left(COOC_{q'}H_{2q'}\right)_2\)
TN2: Đốt hỗn hợp 3 muối.
Đặt \(n_{K_2CO_3}=x;n_{H_2O}=y\left(mol\right)\)
\(BTNT.K\Rightarrow n_{COOK^-}=2n_{K_2CO_3}=2x\left(mol\right)\\ BTNT.O\Rightarrow2n_{COOK^-}+2n_{O_2}=3n_{K_2CO_3}+2n_{CO_2}+n_{H_2O}\\ \Rightarrow x-y=0,3\\ BTKL\Rightarrow m_{M'}+m_{O_2}=m_{K_2CO_3}+m_{CO_2}+m_{H_2O}\\ \Rightarrow138x+18y=99,9\\ \Rightarrow\left\{{}\begin{matrix}x=0,675\\y=0,375\end{matrix}\right.\)
H2 muối gồm: \(C_mH_{2m+1}COOK\text{ }a\text{ }mol;C_nH_{2n-1}COOK\text{ }b\text{ }mol;C_qH_{2q}\left(COOK\right)_2\text{ }c\text{ }mol\)
\(\Rightarrow n_A=a+b+c=0,85\\ BTNT.C\Rightarrow\left(m+1\right)a+\left(n+1\right)b+\left(q+2\right)c=n_{K_2CO_3}+n_{CO_2}=1,75\\ \Rightarrow ma+nb+qc=0,4\\ BTNT.K\Rightarrow a+b+2c=1,35\\ BTNT.H\Rightarrow\left(2m+1\right)a+\left(2n-1\right)b+2qc=2n_{H_2O}=0,75\\ \Rightarrow a-b=-0,05\\ \Rightarrow\left\{{}\begin{matrix}a=0,15\\b=0,2\\c=0,5\end{matrix}\right.\\ \Rightarrow0,15m+0,2n+0,5q=0,4\)
Do \(m;q\ge0\Rightarrow n\le\frac{0,4}{0,2}=2\)
Mà \(n\ge2\Rightarrow n=2\Rightarrow m=q=0\)
\(\text{c) }y=2sin^2x+4\sqrt{3}sinx\cdot cosx+6cos^2x+1\\ =\left(1-cos2x\right)+2\sqrt{3}sin2x+3\left(cos2x+1\right)+1\\ =2cos2x+2\sqrt{3}sin2x+5\)
Đặt \(t=2cos2x+2\sqrt{3}sin2x\)
\(\Rightarrow t^2\le\left[2^2+\left(2\sqrt{3}\right)^2\right]\left(cos^22x+sin^22x\right)=16\\ \Rightarrow-4\le t\le4\\ \Rightarrow1\le y\le9\\ \)
Vậy \(Min\text{ }y=1\Leftrightarrow sin2x=-\frac{1}{2}\)
\(Max\text{ }y=9\Leftrightarrow sin2x=\frac{1}{2}\)
Đặt ẩn rồi xét sự biến thiên thôi
\(\cos x=t;t\in\left[-1;1\right]\)
\(\Rightarrow y=t^2+t-1\)
\(f\left(-1\right)=\left(-1\right)^2-1-1=-1\)
\(f\left(1\right)=1+1-1=1\)
\(f\left(-\frac{1}{2}\right)=\left(-\frac{1}{2}\right)^2-\frac{1}{2}-1=-\frac{5}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}y_{max}=f\left(1\right)=1\\y_{min}=f\left(-\frac{1}{2}\right)=-\frac{5}{4}\end{matrix}\right.\)
Tự xét dấu bằng nhó
