a, \(cos\left(x-\dfrac{\pi}{3}\right)-sin\left(x-\dfrac{\pi}{3}\right)=1\)

\(\Leftrightarrow\sqrt{2}cos\left(x-\dfrac{\pi}{3}-\dfrac{\pi}{4}\right)=1\)

\(\Leftrightarrow cos\left(x-\dfrac{7\pi}{12}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow x-\dfrac{7\pi}{12}=\pm\dfrac{\pi}{4}+k2\pi\)

...

b, \(\sqrt{3}sin2x+2cos^2x=2sinx+1\)

\(\Leftrightarrow\sqrt{3}sin2x+2cos^2x-1=2sinx\)

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

\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{6}\right)=sinx\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)

a: Khi m=1 thì phương trình sẽ là x^2-2x-1=0

=>x^2-2x+1-2=0

=>(x-1)^2=2

=>\(x=\pm\sqrt{2}+1\)

b: Δ=(-2)^2-4*1*(-m^2)=4m^2+4>=4>0

=>Phương trình luôn có hai nghiệm phân biệt

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

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

\(\Leftrightarrow\dfrac{cosx-sinx}{sinx}=cosx\left(cosx-sinx\right)-sinx\left(cosx-sinx\right)\)

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

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-sin2x-cos2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right)=0\)

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(ĐKXĐ:x\in R\)

Phương trình cho tương đương :

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

Đặt \(\sqrt{x^2+1}=a\left(a\ge1\right)\Rightarrow a^2-2=x^2-1\)

Khi đó pt trở thành :

\(a^2\left(a^2-2\right)+a^3=0\)

\(\Leftrightarrow a^2\left(a^2-2+a\right)=0\)

\(\Leftrightarrow a^2\left(a+2\right)\left(a-1\right)=0\)

\(\Leftrightarrow a=1\) ( do \(a\ge1\) )

\(\Rightarrow\sqrt{x^2+1}=1\Rightarrow x^2+1=1\Rightarrow x=0\) ( Thỏa mãn )

Vậy \(S=\left\{0\right\}\)

\(\frac{2\cos2x}{1-\sin2x}=0\Leftrightarrow\hept{\begin{cases}\cos2x=0\\1-\sin2x\ne0\end{cases}}\)

\(\cos2x=0\Leftrightarrow2x\pm\frac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\pm\frac{\pi}{4}+k\pi\)

Với \(x=\frac{\pi}{4}+k\pi\Rightarrow2x=\frac{\pi}{2}+k2\pi\)

\(\sin2x=\sin\left(\frac{\pi}{2}+k2\pi\right)=1\) vi phạm điều kiện \(1-\sin2x\ne0\)

Do đó ta loại nghiệm \(x=\frac{\pi}{4}+k\pi\) của phương trình cos2x = 0

Vậy \(\frac{2\cos2x}{1-\sin2x}=0\Leftrightarrow x=-\frac{\pi}{4}+k\pi,k\in Z\)

Thanks Kikyo =) Nhưng t ko tíc được sory

\(\frac{2\cos2x}{1-\sin2x}=0\)

\(\Leftrightarrow\frac{2\left(\cos^2x-\sin^2x\right)}{\left(\cos x-\sin x\right)^2}=0\)

\(\Leftrightarrow\frac{2\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)}{\left(\cos x-\sin x\right)\left(\cos x-\sin x\right)}=0\)

\(\Leftrightarrow\frac{2\left(\cos x+\sin x\right)}{\cos x-\sin x}=0\rightarrow\text{ĐK: }x\ne\frac{\pi}{4}+k\pi\)

\(\Leftrightarrow\cos x+\sin x=0\)

\(\Leftrightarrow\sqrt{z}\sin\left(x+\frac{\pi}{a}\right)=0\)

\(\Leftrightarrow x+\frac{\pi}{4}=k\pi\)

\(\Leftrightarrow x=\frac{-\pi}{4}+k\pi\)

Vì \(x_1\) là nghiệm PT nên \(x_1^2+3x_1-7=0\Leftrightarrow x_1^2=7-3x_1\)

\(F=x_1^2-3x_2-2013=7-3x_1-3x_2-2013\\ F=-3\left(x_1+x_2\right)-2006\)

Mà theo Viét ta có \(x_1+x_2=-3\)

\(\Rightarrow F=\left(-3\right)\left(-3\right)-2006=-1997\)

ĐKXĐ: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-1}=a\ge0\\\sqrt[3]{2-x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^3=1\)

Ta được hệ: 

\(\left\{{}\begin{matrix}a+b=1\\a^2+b^3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\a^2+b^3=1\end{matrix}\right.\)

\(\Rightarrow a^2+\left(1-a\right)^3=1\)

\(\Leftrightarrow a^3-4a^2+3a=0\)

\(\Leftrightarrow a\left(a-1\right)\left(a-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x-1}=0\\\sqrt[]{x-1}=1\\\sqrt[]{x-1}=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=10\end{matrix}\right.\)