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\(\Leftrightarrow\dfrac{u_{n+1}}{n+1}=\dfrac{1}{3}.\dfrac{u_n}{n}\)

Đặt \(\dfrac{u_n}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{1}{3}\\v_{n+1}=\dfrac{1}{3}v_n\end{matrix}\right.\)

\(\Rightarrow v_n\) là CSN với công bội \(\dfrac{1}{3}\)

\(\Rightarrow v_n=\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^{n-1}=\left(\dfrac{1}{3}\right)^n\)

\(S=\sum\limits^{10}_{k=1}\left(\dfrac{1}{3}\right)^k=\dfrac{\dfrac{1}{3}\left(1-\dfrac{1}{3^{10}}\right)}{1-\dfrac{1}{3}}=\dfrac{1}{2}\left(1-\dfrac{1}{3^{10}}\right)\)

\(\hept{\begin{cases}y=2\sqrt{x-1}\left(1\right)\\\sqrt{x+y}=x^2-y\left(2\right)\end{cases}}\)  (ĐKXĐ: \(x\ge1;x\ge-y;\left(x;y\right)\in R\))

Thế (1) vào (2) ta được phương trình: \(\sqrt{x+2\sqrt{x-1}}=x^2-2\sqrt{x-1}\)

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

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

\(\Leftrightarrow\sqrt{x-1}+1=x^2-2\sqrt{x-1}\) (Do \(\sqrt{x-1}+1>0\))

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-3\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}\left(x+1\right)-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\\sqrt{x-1}\left(x+1\right)=3\left(3\right)\end{cases}}\)

\(\left(3\right)\Leftrightarrow x^3+x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(x^2+3x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^2+3x+5=0\left(vn\right)\end{cases}\Leftrightarrow}x=2\). Từ (1) suy ra: \(y=2\)

Vậy hệ PT cho có nghiệm duy nhất (x;y)=(2;2)

Bổ sung: Với x=1, từ (1) suy ra y=0 => (x;y)=(1;0)

a/ \(f'\left(x\right)=12sin^33x.cos3x\)

\(f'\left(x\right)=g\left(x\right)\Leftrightarrow12sin^33x.cos3x=sin6x\)

\(\Leftrightarrow6sin^23x.2sin3x.cos3x-sin6x=0\)

\(\Leftrightarrow6sin^23x.sin6x-sin6x=0\)

\(\Leftrightarrow sin6x\left(6sin^23x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}sin6x=0\\sin^23x=\frac{1}{6}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\\\frac{1-cos6x}{2}=\frac{1}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin6x=0\\cos6x=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x=k\pi\\6x=a+k2\pi\\6x=-a+k2\pi\end{matrix}\right.\) với \(cosa=\frac{2}{3}\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{k\pi}{6}\\x=\frac{a}{6}+\frac{k\pi}{3}\\x=-\frac{a}{6}+\frac{k\pi}{3}\end{matrix}\right.\)

b/

\(f'\left(x\right)=6sin^22x.cos2x=4cos2x-5sin4x\)

\(\Leftrightarrow6sin^22x.cos2x=4cos2x-10sin2x.cos2x\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\3sin^22x=2-5sin2x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow3sin^22x+5sin2x-2=0\)

\(\Rightarrow\left[{}\begin{matrix}sin2x=\frac{1}{3}\\sin2x=-2< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow sin2x=sina\) (với \(sina=\frac{1}{3}\))

\(\Rightarrow\left[{}\begin{matrix}2x=a+k2\pi\\2x=\pi-a+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{a}{2}+k\pi\\x=\frac{\pi}{2}-\frac{a}{2}+k\pi\end{matrix}\right.\)

c/

\(f'\left(x\right)=4x.cos^2\frac{x}{2}-2x^2.cos\frac{x}{2}.sin\frac{x}{2}=2x\left(1+cosx\right)-x^2sinx\)

\(f'\left(x\right)=g\left(x\right)\)

\(\Leftrightarrow2x\left(1+cosx\right)-x^2sinx=x-x^2sinx\)

\(\Leftrightarrow2x\left(1+cosx\right)=x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2\left(1+cosx\right)=1\left(1\right)\end{matrix}\right.\)

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

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