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1.
Trước hết bạn nhớ công thức:
$1^2+2^2+....+n^2=\frac{n(n+1)(2n+1)}{6}$ (cách cm ở đây: https://hoc24.vn/cau-hoi/tinh-tongs-122232n2.83618073020)
Áp vào bài:
\(\lim\frac{1}{n^3}[1^2+2^2+....+(n-1)^2]=\lim \frac{1}{n^3}.\frac{(n-1)n(2n-1)}{6}=\lim \frac{n(n-1)(2n-1)}{6n^3}\)
\(=\lim \frac{(n-1)(2n-1)}{6n^2}=\lim (\frac{n-1}{n}.\frac{2n-1}{6n})=\lim (1-\frac{1}{n})(\frac{1}{3}-\frac{1}{6n})\)
\(=1.\frac{1}{3}=\frac{1}{3}\)
2.
\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)
\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)
\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)
\(=x+a\)
a) \(\left(\dfrac{1}{2}\right)^n\le10^{-9}\)\(\Leftrightarrow2^{-n}\le10^{-9}\)\(\Leftrightarrow-n\le log^{10^{-9}}_2\)\(\Leftrightarrow-n\le-9log^{10}_2\)\(\Leftrightarrow n\ge9log^{10}_2\)\(\Leftrightarrow n\ge30\).
Vậy \(n=30\).
b) \(3-\left(\dfrac{7}{5}\right)^n\le0\)
\(\Leftrightarrow-\left(\dfrac{7}{5}\right)^n\le-3\)
\(\Leftrightarrow\left(\dfrac{7}{5}\right)^n\ge3\)\(\Leftrightarrow n\ge log^3_{\dfrac{7}{5}}\)
\(\Rightarrow\)\(n\in\left\{4;5;6;7;...\right\}\Rightarrow n=4\)
c) \(1-\left(\dfrac{4}{5}\right)^n\ge0,97\)
\(\Leftrightarrow-\left(\dfrac{4}{5}\right)^n\ge-0,3\)
\(\Leftrightarrow\left(\dfrac{4}{5}\right)^n\le0,3\)\(\Leftrightarrow n\ge log^{0,3}_{\dfrac{4}{5}}\)
\(\Rightarrow n\in\left\{6;7;8;9...\right\}\Rightarrow n=6\)
d)\(\left(1+\dfrac{5}{100}\right)^n\ge2\)
\(\Leftrightarrow1,05^n\ge2\)
\(\Rightarrow n\in\left\{15;16;17;18;...\right\}\Rightarrow n=15\)
\(y=x+\dfrac{1}{x}-5\ge2\sqrt{\dfrac{x}{x}}-5=-3\)
\(y_{min}=-3\) khi \(x=1\)
\(y=4x^2+\dfrac{1}{2x}+\dfrac{1}{2x}-4\ge3\sqrt[3]{\dfrac{4x^2}{2x.2x}}-4=-1\)
\(y_{min}=-1\) khi \(x=\dfrac{1}{2}\)
\(y=x+\dfrac{4}{x}\Rightarrow y'=1-\dfrac{4}{x^2}=0\Rightarrow x=-2\)
\(y\left(-2\right)=-4\Rightarrow\max\limits_{x>0}y=-4\) khi \(x=-2\)
\(\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\dfrac{x^2+\left(x+1\right)^2+x^2\left(x+1\right)^2}{x^2\left(x+1\right)^2}}=\sqrt{\dfrac{x^2\left(x+1\right)^2+2x^2+2x+1}{x^2\left(x+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(x^2+x\right)^2+2\left(x^2+x\right)+1}{\left(x^2+x\right)^2}}=\sqrt{\dfrac{\left(x^2+x+1\right)^2}{\left(x^2+x\right)^2}}=\dfrac{x^2+x+1}{x^2+x}\)
\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)
\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)
\(=5^{2021-\dfrac{1}{2021}}\)
\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)
\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)
Biến đổi: ʃ\(\int\dfrac{1dx}{cosx\dfrac{\sqrt{2}}{2}\left(cosx-sinx\right)}=\int\dfrac{\sqrt{2}dx}{cos^2x\left(1-tanx\right)}=\int\dfrac{\sqrt{2}d\left(tanx\right)}{1-tanx}=-\sqrt{2}\ln trituyetdoi\left(1-tanx\right)\)
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Lời giải:
\(\frac{3x^3f(x)}{f'(x)^2+xf'(x)+x^2}=f'(x)-x\)
\(\Rightarrow 3x^3f(x)=[f'(x)-x][f'(x)^2+xf'(x)+x^2]=f'(x)^3-x^3\)
\(\Rightarrow 3f(x)=\left(\frac{f'(x)}{x}\right)^3-1\)
Đặt \(\frac{f'(x)}{x}=g(x)\Rightarrow f'(x)=xg(x)(1)\) .
Vì \(f(1)=\frac{7}{3}\Rightarrow f'(1)=2\Rightarrow g(1)=2\)
Ta có: \(3f(x)=g(x)^3-1\)
\(\Rightarrow 3f'(x)=3g'(x)g(x)^2\)
\(\Rightarrow f'(x)=g'(x)g(x)^2(2)\)
Từ \((1);(2)\Rightarrow xg(x)=g'(x)g(x)^2\)
\(\Rightarrow x=g'(x)g(x)=\frac{1}{2}[g(x)^2]'\) \(\Rightarrow 2x=[g(x)^2]'\Rightarrow g(x)^2=\int 2xdx=x^2+c\)
Kết hợp với $g(1)=2$ suy ra $c=3$
Vậy \(g(x)^2=x^2+3\Rightarrow f(x)=\frac{g(x)^3-1}{3}=\frac{(x^2+3)^{\frac{3}{2}}-1}{3}\)
\(\Rightarrow f(2)=\frac{\sqrt{343}-1}{3}\)
Bài đã đăng bạn hạn chế không đăng lại nữa nhé.