\(lim\left(\sqrt{4n^2+2}\sqrt[3]{n^3+1}-2n\sqrt[3]{n^3+2}\right)\\ =lim\left[\sqrt[3]{n^3+1}\left(\sqrt{4n^2+2}-2n\right)-2n\left(\sqrt[3]{n^3+2}-\sqrt[3]{n^3+1}\right)\right]\\ =lim\left[\dfrac{2\sqrt[3]{n^3+1}}{\sqrt{4n^2+2}+2n}-\dfrac{2n}{\left(\sqrt[3]{n^3+2}\right)^2+\left(\sqrt[3]{n^3+1}\right)^2+\sqrt[3]{n^3+2}\sqrt[3]{n^3+1}}\right]\\ =lim\left[\dfrac{2\sqrt[3]{1+\dfrac{1}{n^3}}}{\sqrt{4+\dfrac{2}{n^2}}+2}-\dfrac{\dfrac{2}{n}}{\left(\sqrt[3]{1+\dfrac{2}{n}}\right)^2+\left(\sqrt[3]{1+\dfrac{1}{n}}\right)^2+\sqrt[3]{1+\dfrac{2}{n}}\sqrt[3]{1+\dfrac{1}{n}}}\right]\)

\(=\dfrac{1}{2}\)

@Nguyễn Mạnh Vũ: \(9x^2-6x+1=\left(3x-1\right)^2\ge0\) nên \(6x\le9x^2+1\)

\(\left(x+3\right)^2=x^2+6x+9\le x^2+\left(9x^2+1\right)+9=10\left(x^2+1\right)\)

Suy ra: \(P=\dfrac{x+3}{\sqrt{x^2+1}}\le\sqrt{10}\)

Vậy \(MaxP=\sqrt{10}\) (khi \(x=\dfrac{1}{3}\))

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

TH1: \(cosx=0\)

Phương trình có nghiệm \(x=\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

TH2: \(cosx\ne0\)

Phương trình tương đương: \(\left(tanx+1\right)^3-3\left(1+tan^2x\right)-tanx\left(1+tan^2x\right)=0\\ \Leftrightarrow tanx=1\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

\(VT=\left(cos2x-cos4x\right)^2=\left(-2cos^22x+cos2x+1\right)^2\le\left(\dfrac{9}{8}\right)^2\\ VP=6+2sin3x\ge4\\ \rightarrow VT< VP\)

Suy ra phương trình vô nghiệm.

Đặt \(v_n=u_n+n\)

Chứng minh được \(3^n>n^2\) với mọi số nguyên dương n bằng phương pháp quy nạp. Suy ra: \(\left|\dfrac{n}{3^n}\right|< \left|\dfrac{n}{n^2}\right|=\dfrac{1}{n}\). Mà \(lim\dfrac{1}{n}=0\rightarrow lim\dfrac{n}{3^n}=0\)

\(u_{n+1}=3u_n+2n-1\rightarrow v_{n+1}=3v_n\\ \rightarrow v_n=v_1.3^{n-1}=2.3^{n-1}\\ \rightarrow u_n=2.3^{n-1}-n\\ lim\dfrac{u_n}{3^n}=lim\dfrac{2.3^{n-1}-n}{3^n}=lim\left(\dfrac{2}{3}-\dfrac{n}{3^n}\right)=\dfrac{2}{3}\)

Quá sai.

Sai đoạn cuối.

\(sin^23xcos2x+sin^2x=0\rightarrow\dfrac{1-cos6x}{2}.cos2x+\dfrac{1-cos2x}{2}=0\\ \rightarrow cos6xcos2x=1\rightarrow cos8x+cos4x=2\\ \rightarrow cos8x=cos4x=1\rightarrow x=\dfrac{k\pi}{2}\left(k\in Z\right)\)

Sai.