2. ĐKXĐ:

a. \(\left\{{}\begin{matrix}cosx\ne0\\2-cosx+tan^2x\ge0\left(luôn-đúng\right)\end{matrix}\right.\)

\(\Rightarrow x\ne\frac{\pi}{2}+k\pi\)

(BPT dưới luôn đúng do \(\left\{{}\begin{matrix}tan^2x\ge0\\2-cosx>0\end{matrix}\right.\) với mọi x)

b. \(sin2x-sinx+3\ge0\)

\(\Leftrightarrow\left(sin2x+2\right)+\left(1-sinx\right)\ge0\)

Do \(\left\{{}\begin{matrix}sin2x\ge-1\\sinx\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}sin2x+2>0\\1-sinx\ge0\end{matrix}\right.\)

\(\Rightarrow\) BPT luôn thỏa mãn hay hàm số xác định trên R

1.

\(\Leftrightarrow f\left(x\right)=sin^4x+cos^4x-2m.sinx.cosx\ge0\) ;\(\forall x\in R\)

\(f\left(x\right)=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x-2m.sinx.cosx\)

\(=-\frac{1}{2}sin^22x-m.sin2x+1\)

Đặt \(sin2x=t\Rightarrow\left|t\right|\le1\)

\(f\left(t\right)=-\frac{1}{2}t^2-mt+1\ge0\) ; \(\forall t\in\left[-1;1\right]\)

\(\Leftrightarrow\min\limits_{\left[-1;1\right]}f\left(t\right)\ge0\)

\(a=-\frac{1}{2}< 0\Rightarrow\min\limits f\left(t\right)\) xảy ra tại 1 trong 2 đầu mút

\(f\left(-1\right)=m+\frac{1}{2}\) ; \(f\left(1\right)=\frac{1}{2}-m\)

TH1: \(\left\{{}\begin{matrix}m+\frac{1}{2}\ge\frac{1}{2}-m\\\frac{1}{2}-m\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\ge0\\m\le\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow0\le m\le\frac{1}{2}\)

TH2: \(\left\{{}\begin{matrix}\frac{1}{2}-m\ge m+\frac{1}{2}\\m+\frac{1}{2}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\le0\\m\ge-\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow-\frac{1}{2}\le m\le\frac{1}{2}\)

Lời giải:ĐKXĐ: \(\left\{\begin{matrix} 6-x\geq 0\\ x-1\geq 0\\ 1+\sqrt{x-1}\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 6\\ x\geq 1\end{matrix}\right.\) hay $x\in [1;6]$ 

Đáp án D

ĐKXĐ: \(\begin{cases}m-2x\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}2x\le m\\ x\ge-1\end{cases}\)

=>\(\begin{cases}x\le\frac{m}{2}\\ x\ge-1\end{cases}\)

=>-1<=x<=m/2

Để tập xác định là một đoạn trên trục số (tức là tập xác định có độ dài lớn hơn 0, hay đoạn đó không bị suy biến thành một điểm hoặc tập rỗng), ta cần điều kiện là -1<m/2

=>-2<m

=>m>-2

=>Chọn D

\(\left\{{}\begin{matrix}\sqrt{x-2\sqrt{x-1}}\ne0\\x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\\left(\sqrt{x-1}-1\right)^2\ge0\\x\ge1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\in R\\x\ge1\end{matrix}\right.\)

\(\Rightarrow TXĐ:D=[1;+\infty)\cup\left\{2\right\}\)

1. \(sin\left(\dfrac{\pi}{3}-x\right)\ne0\Leftrightarrow\dfrac{\pi}{3}-x\ne k\pi\Leftrightarrow x\ne\dfrac{\pi}{3}-k\pi\)

2. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

3. \(\sqrt{1+sinx}-\sqrt{2}\ge0\Leftrightarrow1+sinx\ge2\Leftrightarrow sinx\ge1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

4. \(\sqrt{2-2cosx}-2\ne0\Leftrightarrow2-2cosx\ne4\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)

5. \(1-\sqrt{1+sin3x}\ne0\Leftrightarrow sin3x\ne0\Leftrightarrow3x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{3}\)

câu 4 sao ra 2-2cosx\(\ne\)4 ạ

2.1

a.

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

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

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

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

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

b.

\(cosx-\sqrt{3}sinx=1\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)

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

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

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

a: ĐKXĐ: \(\sqrt3\cdot\sin x+cosx<>0\)

=>\(\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx<>0\)

=>\(\sin\left(x+\frac{\pi}{6}\right)<>0\)

=>\(x+\frac{\pi}{6}<>k\pi\)

=>\(x<>-\frac{\pi}{6}+k\pi\)

\(\frac{2\cdot cos2x+1}{\sqrt3\cdot\sin x+cosx}=2\cdot cosx-1\)

=>\(\frac{2\cdot\left(2\cdot cos^2x-1\right)+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}=2\cdot cosx-1\)

=>\(\frac{4\cdot cos^2x-1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-\left(2\cdot cosx-1\right)=0\)

=>\(\left(2\cdot cosx-1\right)\left\lbrack\frac{2\cdot cosx+1}{2\cdot\sin\left(x+\frac{\pi}{6}\right)}-1\right\rbrack=0\)

TH1: \(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}-1=0\)

=>\(\frac{2\cdot cosx+1}{\sqrt3\cdot\sin x+cosx}=1\)

=>\(\sqrt3\cdot\sin x+cosx=2\cdot cosx+1\)

=>\(\sqrt3\cdot\sin x-cosx=1\)

=>\(\frac{\sqrt3}{2}\cdot\sin x-\frac12\cdot cosx=\frac12\)

=>\(\sin\left(x-\frac{\pi}{6}\right)=\frac12\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{6}=\frac{\pi}{6}+k2\pi\\ x-\frac{\pi}{6}=\pi-\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=\pi+k2\pi\end{array}\right.\)

TH2: \(2\cdot cosx-1=0\)

=>\(cosx=\frac12\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{3}+k2\pi\\ x=-\frac{\pi}{3}+k2\pi\end{array}\right.\)