Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:

\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)

\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)

\(=-1\)

\(32sin^6\dfrac{x}{2}+sin3x=3sinx\)

\(\Leftrightarrow32sin^6\dfrac{x}{2}+3sinx-4sin^3x=3sinx\)

\(\Leftrightarrow8sin^6\dfrac{x}{2}=sin^3x\)

\(\Leftrightarrow8sin^6\dfrac{x}{2}=8sin^3\dfrac{x}{2}.cos^3\dfrac{x}{2}\)

\(\Leftrightarrow sin^3\dfrac{x}{2}\left(1-cos^3\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{x}{2}=0\\cos\dfrac{x}{2}=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=k4\pi\end{matrix}\right.\)

\(\Leftrightarrow x=k2\pi\)

\(\lim\limits_{x\rightarrow1}\dfrac{x^3+3x^2-4}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x^2+4x+4\right)}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\left(x+2\right)^2\)

\(=9\)

b, Phương trình hoành độ giao điểm:

\(\Rightarrow2x^2=x+3\)

\(\Leftrightarrow2x^2-x-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(x=-1\Rightarrow y=2\Rightarrow\left(-1;2\right)\)

\(x=\dfrac{3}{2}\Rightarrow y=\dfrac{9}{2}\Rightarrow\left(\dfrac{3}{2};\dfrac{9}{2}\right)\)

a, Phương trình đường thẳng AB có dạng \(y=ax+b\).

\(\Rightarrow\left\{{}\begin{matrix}b=3\\2=b-a\end{matrix}\right.\)

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

\(\Rightarrow y=x+3\)

c, \(P\in Z\Leftrightarrow\dfrac{x}{x+1}\in Z\)

\(\Leftrightarrow1-\dfrac{1}{x+1}\in Z\)

\(\Leftrightarrow x+1\in\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{-2;0\right\}\)

b, \(P=\dfrac{x}{x+1}=\dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+1}=\dfrac{\dfrac{1}{3}}{\dfrac{4}{3}}=\dfrac{1}{4}\)

a, ĐK: \(x\ne\pm1\)

\(P=\left(\dfrac{x+1}{x-1}-\dfrac{4x^2}{1-x^2}-\dfrac{x-1}{x+1}\right).\dfrac{x^2-2x+1}{4x^2-4}\)

\(=\left[\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right].\dfrac{\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x^2+2x+1+4x^2-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\dfrac{x-1}{4\left(x+1\right)}\)

\(=\dfrac{4x^2+4x}{x+1}.\dfrac{1}{4\left(x+1\right)}\)

\(=\dfrac{4x\left(x+1\right)}{x+1}.\dfrac{1}{4\left(x+1\right)}\)

\(=\dfrac{x}{x+1}\)