\(\times\frac{\ge\sin\cos\tan\widehat{\widehat{\stackrel{\frown}{\sum_{\sum_{\prod_{\prod_{\lim_{\int\begin{cases}\left[\begin{matrix}\\\\\end{matrix}\right.\\\end{cases}}}^{ }}^{ }}^{ }}^{ }}}}}{ }\)
😆 😆 😆 😆 😆 😆 😆 😆 😆 😆 😍 😘 😎 😉 🤗 😗 ☺️ 😐 🤗 \(\times\frac{\ge\times_{^{22}\sqrt{\sqrt{\ln\ln\ln\ln\ln\ln\cot\tan\cos\sum_{\prod_{\int\begin{cases}\prod_{\begin{cases}\begin{cases}\begin{cases}\begin{cases}\left[\begin{matrix}\lambda\infty\partial\\\\\end{matrix}\right.\\\end{cases}\\\end{cases}\\\end{cases}\\\end{cases}}^{ }\\\end{cases}}^{ }}^{ }}}}}{ }\)
cac ki hieu trong toan hoc \(\sqrt{\sqrt[]{}\frac{ }{ }\hept{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}^{ }^2_{ }|^{ }_{ }\widehat{ }\widebat{ }\cos\tanh\cosh\sinh\tan\sin\cot\Leftrightarrow\Rightarrow\Leftarrow\rightarrow\leftarrow\leftrightarrow\uparrow\downarrow\ne=\subset\supset\notin\exists\le\ge\forall⋮⊥\subseteq\supseteq\approx\in\lambda\pi\Phi\mu}\)
2\(\sum_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\prod_{\lambda\theta\rho}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }}^{ }\)
cho dãy số (un) \(\left\{{}\begin{matrix}u_1=\dfrac{1}{3}\\u_{n+1}=\dfrac{n+1}{3n}.u_n,n\ge1\end{matrix}\right.\)tính tổng S=\(\sum_{k=1}^{10}\)\(\dfrac{u_k}{k}\)?
\(\Leftrightarrow\dfrac{u_{n+1}}{n+1}=\dfrac{1}{3}.\dfrac{u_n}{n}\)
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 \(\dfrac{1}{3}\)
\(\Rightarrow v_n=\dfrac{1}{3}.\left(\dfrac{1}{3}\right)^{n-1}=\left(\dfrac{1}{3}\right)^n\)
\(S=\sum\limits^{10}_{k=1}\left(\dfrac{1}{3}\right)^k=\dfrac{\dfrac{1}{3}\left(1-\dfrac{1}{3^{10}}\right)}{1-\dfrac{1}{3}}=\dfrac{1}{2}\left(1-\dfrac{1}{3^{10}}\right)\)
Giai hệ phương trình:
\(\hept{\begin{cases}2x^3y+3x^2=5y\\1+6xy=7y^2\end{cases}}\)
\(\left\{{}\begin{matrix}2x^3y+3x^2=5y\\1+6xy=7y^2\end{matrix}\right.\)
Giai hệ phương trình:
\(\hept{\begin{cases}y=2\sqrt{x-1}\\\sqrt{x+y}=x^2-y\end{cases}}\)\(\left\{{}\begin{matrix}y=2\sqrt{x-1}\\\sqrt{x+y}=x^2-y\end{matrix}\right.\)
\(\hept{\begin{cases}y=2\sqrt{x-1}\left(1\right)\\\sqrt{x+y}=x^2-y\left(2\right)\end{cases}}\) (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)
Giải phương trình f'(x) = g(x) với
a) \(\left\{{}\begin{matrix}f\left(x\right)=sin^43x\\g\left(x\right)=sin6x\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}f\left(x\right)=sin^32x\\g\left(x\right)=4cos2x-5sin4x\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}f\left(x\right)=2x^2cos^2\frac{x}{2}\\g\left(x\right)=x-x^2sinx\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}f\left(x\right)=4xcos^2\frac{x}{2}\\g\left(x\right)=8cos\frac{x}{2}-3-2sinx\end{matrix}\right.\)
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/
\(f'\left(x\right)=4x.cos^2\frac{x}{2}-2x^2.cos\frac{x}{2}.sin\frac{x}{2}=2x\left(1+cosx\right)-x^2sinx\)
\(f'\left(x\right)=g\left(x\right)\)
\(\Leftrightarrow2x\left(1+cosx\right)-x^2sinx=x-x^2sinx\)
\(\Leftrightarrow2x\left(1+cosx\right)=x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2\left(1+cosx\right)=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow cosx=-\frac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2\pi}{3}+k2\pi\\x=-\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
Giải hệ phương trình: \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x+2y^3=x^2y\end{matrix}\right.\)
\(\hept{\begin{cases}x^2+xy+y^2=7\\x+2y^3=x^2y\end{cases}}\)
Giải hệ PT: \(\hept{\begin{cases}x^2-y^2=1-xy\\x^2+y^2=3xy+11\end{cases}}\)\(\left\{{}\begin{matrix}x^2-y^2=1-xy\\x^2+y^2=3xy+11\end{matrix}\right.\)