ta có 

\(P=sin8x-2sinxcos7x-2sinxcos5x=sin8x-\left(sin8x-sin6x\right)-\left(sin6x-sin4x\right)\)

\(=sin4x\)

\(cos\left(x\right)-cos\left(2x\right)=sin\left(3x\right)\)

\(\Leftrightarrow-2sin\frac{3x}{2}sin\frac{-x}{2}=2sin\frac{3x}{2}cos\frac{3x}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}sin\frac{3x}{2}=0\left(1\right)\\sin\frac{x}{2}=cos\frac{3x}{2}\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\frac{3x}{2}=k\pi\left(k\inℤ\right)\)

\(\Leftrightarrow x=\frac{2k\pi}{3}\left(k\inℤ\right)\)

\(\left(2\right)\Leftrightarrow sin\frac{x}{2}=sin\left(\frac{\pi}{2}-\frac{3x}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{x}{2}=\frac{\pi}{2}-\frac{3x}{2}+k2\pi\\\frac{x}{2}=\pi-\left(\frac{\pi}{2}-\frac{3x}{2}\right)+k2\pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{4}+k\pi\\x=-\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)

Gọi \(M\left(0,y\right)\in Oy\)

ta có M cách đều A,B hay \(MA=MB\Leftrightarrow1+y^2=2^2+\left(y-3\right)^2\)

\(\Leftrightarrow6y=12\Leftrightarrow y=2\)

Vậy tọa độ của M khi đó là (0,2)

a.\(\left(3-x\right)\left(x^2+5x+6\right)=\left(3-x\right)\left(x+2\right)\left(x+3\right)\)

ta có : 

Vậy bất phương trình có nghiệm \(\text{(}-\infty,-3\text{]}\cup\left[-2,3\right]\)

b. \(\left(6+5x\right)\left(x^2-5x+6\right)=\left(6+5x\right)\left(x-2\right)\left(x-3\right)\)

Vậu BPT có nghiệm \(\left[-\frac{6}{5},2\right]\cup\text{[}3,+\infty\text{)}\)

\(a)\)

\(1-sin\left(x\right)\)

\(=sin^2\frac{x}{2}+cos^2\frac{x}{2}-2.sin\frac{x}{2}.cos\frac{x}{2}\)

\(=\left(sin\frac{x}{2}-cos\frac{x}{2}\right)^2\)

\(b)\)

\(1+sin\left(x\right)\)

\(=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2.sin\frac{x}{2}.cos\frac{x}{2}\)

\(=\left(sin\frac{x}{2}+cos\frac{x}{2}\right)^2\)

\(d)\)

\(1+2cos\left(x\right)\)

\(=2\left(\frac{1}{2}+cosx\right)\)

\(=2\left(cos60^o+cosx\right)\)

\(=4\left(cos\frac{60^o+x}{2}cos\frac{60^o-x}{2}\right)\)

\(=4cos\left(30^o+\frac{x}{2}\right)cos\left(30^o-\frac{x}{2}\right)\)

ta có điều kiện \(\hept{\begin{cases}2x-1\ge0\\x-1\ge0\\3x+1\ne2x-1\end{cases}\Leftrightarrow x\ge1}\)

vậy BPT \(\Leftrightarrow\frac{3x\left(\sqrt{2x-1}-\sqrt{x-1}\right)}{2x-1-\left(x-1\right)}-\frac{\left(x+2\right)\left(\sqrt{3x+1}+\sqrt{2x-1}\right)}{3x+1-\left(2x-1\right)}>0\)

\(\Leftrightarrow3\left(\sqrt{2x-1}-\sqrt{x-1}\right)-\left(\sqrt{3x+1}+\sqrt{2x-1}\right)>0\)

\(\Leftrightarrow2\sqrt{2x-1}>\sqrt{3x+1}+3\sqrt{x-1}\Leftrightarrow8x-4>12x-8+6\sqrt{3x+1}.\sqrt{x-1}\)

\(\Leftrightarrow4-4x>6\sqrt{3x+1}.3\sqrt{x-1}\) vô lí do vế trái \(\le0\forall x\ge1\) còn vế phải lớn hơn bằng không

vậy bất phương trình vô nghiệm

ko bt

ko bt

ko bt

KO ĂN GIAN KHI HỌC

ko ăn gian