Bạn tham khảo pt 1 hộ mình nha. Chúc bạn học tốt~

Pt 1.

Bạn tham khảo phương trình 1 hộ mình nha. Chúc bạn học tốt

a: ĐKXĐ: 2*sin x+1<>0

=>sin x<>-1/2

=>x<>-pi/6+k2pi và x<>7/6pi+k2pi

b: ĐKXĐ: \(\dfrac{1+cosx}{2-cosx}>=0\)

mà 1+cosx>=0

nên 2-cosx>=0

=>cosx<=2(luôn đúng)

c ĐKXĐ: tan x>0

=>kpi<x<pi/2+kpi

d: ĐKXĐ: \(2\cdot cos\left(x-\dfrac{pi}{4}\right)-1< >0\)

=>cos(x-pi/4)<>1/2

=>x-pi/4<>pi/3+k2pi và x-pi/4<>-pi/3+k2pi

=>x<>7/12pi+k2pi và x<>-pi/12+k2pi

e: ĐKXĐ: x-pi/3<>pi/2+kpi và x+pi/4<>kpi

=>x<>5/6pi+kpi và x<>kpi-pi/4

f: ĐKXĐ: cos^2x-sin^2x<>0

=>cos2x<>0

=>2x<>pi/2+kpi

=>x<>pi/4+kpi/2

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k2\pi\\x\ne-\dfrac{\pi}{6}+k2\pi\\x\ne\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x+sinx}=\sqrt{3}\)

\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x+\sqrt{3}sinx\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

ĐKXĐ : \(sinx\ne1;-\dfrac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+2k\pi\\x\ne\dfrac{-\pi}{6}+2k\pi;\dfrac{7\pi}{6}+2k\pi\end{matrix}\right.\)   

\(\Leftrightarrow x\ne\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\)( k thuộc Z ) 

P/t đã cho \(\Leftrightarrow\dfrac{cosx-sin2x}{1-2sin^2x+sinx}=\sqrt{3}\) 

\(\Leftrightarrow cosx-sin2x=\sqrt{3}\left(cos2x+sinx\right)\)

\(\Leftrightarrow cosx-\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=cos\left(2x+\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=x+\dfrac{\pi}{3}+2k\pi\\2x+\dfrac{\pi}{6}=-x-\dfrac{\pi}{3}+2k\pi\end{matrix}\right.\) ( k thuộc Z ) 

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+2k\pi\\x=\dfrac{-\pi}{6}+\dfrac{2}{3}k\pi\left(L\right)\end{matrix}\right.\)

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