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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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1/ \(I=\int\limits^1_0\dfrac{2x+1}{x^2+x+1}dx=\int\limits^1_0\dfrac{d\left(x^2+x+1\right)}{x^2+x+1}=ln\left|x^2+x+1\right||^1_0=ln3\)
2/ \(\int\limits^{\dfrac{1}{2}}_0\dfrac{5x}{\left(1-x^2\right)^3}dx=-\dfrac{5}{2}\int\limits^{\dfrac{1}{2}}_0\dfrac{d\left(1-x^2\right)}{\left(1-x^2\right)^3}=\dfrac{5}{4}\dfrac{1}{\left(1-x^2\right)^2}|^{\dfrac{1}{2}}_0=\dfrac{35}{36}\)
3/ \(\int\limits^1_0\dfrac{2x}{\left(x+1\right)^3}dx\Rightarrow\) đặt \(x+1=t\Rightarrow x=t-1\Rightarrow dx=dt;\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=1\Rightarrow t=2\end{matrix}\right.\)
\(I=\int\limits^2_1\dfrac{2\left(t-1\right)dt}{t^3}=\int\limits^2_1\left(\dfrac{2}{t^2}-\dfrac{2}{t^3}\right)dt=\left(\dfrac{-2}{t}+\dfrac{1}{t^2}\right)|^2_1=\dfrac{1}{4}\)
4/ \(\int\limits^1_0\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}dx\)
Kĩ thuật chung là tách và sử dụng hệ số bất định như sau:
\(\dfrac{4x-2}{\left(x^2+1\right)\left(x+2\right)}=\dfrac{ax+b}{x^2+1}+\dfrac{c}{x+2}=\dfrac{\left(a+c\right)x^2+\left(2a+b\right)x+2b+c}{\left(x^2+1\right)\left(x+2\right)}\)
\(\Rightarrow\left\{{}\begin{matrix}a+c=0\\2a+b=4\\2b+c=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=0\\a=-c=2\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^1_0\left(\dfrac{2x}{x^2+1}-\dfrac{2}{x+2}\right)dx=\int\limits^1_0\dfrac{d\left(x^2+1\right)}{x^2+1}-2\int\limits^1_0\dfrac{d\left(x+2\right)}{x+2}=ln\dfrac{8}{9}\)
5/ \(\int\limits^1_0\dfrac{x^2dx}{x^6-9}\Rightarrow\) đặt \(x^3=t\Rightarrow3x^2dx=dt\Rightarrow x^2dx=\dfrac{1}{3}dt;\left\{{}\begin{matrix}x=0\Rightarrow t=0\\x=1\Rightarrow t=1\end{matrix}\right.\)
\(I=\dfrac{1}{3}\int\limits^1_0\dfrac{dt}{t^2-9}=\dfrac{1}{18}\int\limits^1_0\left(\dfrac{1}{t-3}-\dfrac{1}{t+3}\right)dt=\dfrac{1}{18}ln\left|\dfrac{t-3}{t+3}\right||^1_0=-\dfrac{1}{18}ln2\)
6/ Tương tự câu 4, sử dụng hệ số bất định ta tách được:
\(\int\limits^2_1\dfrac{2x-1}{x^2\left(x+1\right)}dx=\int\limits^2_1\left(\dfrac{3x-1}{x^2}-\dfrac{3}{x+1}\right)dx=\int\limits^2_1\left(\dfrac{3}{x}-\dfrac{1}{x^2}-\dfrac{3}{x+1}\right)dx\)
\(=\left(3ln\left|\dfrac{x}{x+1}\right|+\dfrac{1}{x}\right)|^2_1=3ln\dfrac{4}{3}-\dfrac{1}{2}\)
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.
\(\lim \frac{1}{n}\left[(x+\frac{a}{n})+(x+\frac{2a}{n})+...+(x.\frac{(n-1)a}{n}\right]\)
\(=\lim \frac{1}{n}\left[\underbrace{(x+x+...+x)}_{n-1}+\frac{a(1+2+...+n-1)}{n} \right]\)
\(=\lim \frac{1}{n}[(n-1)x+a(n-1)]=\lim \frac{n-1}{n}(x+a)=\lim (1-\frac{1}{n})(x+a)\)
\(=x+a\)
1)
Ta có:
\(\int (2-\cot ^2x)dx=\int (2-\frac{\cos ^2x}{\sin ^2x})dx\)
\(=\int (2-\frac{1-\sin ^2x}{\sin ^2x})dx=\int (3-\frac{1}{\sin ^2x})dx=3\int dx-\int \frac{dx}{\sin ^2x}\)
\(=3x+\int d(\cot x)=3x+\cot x+c\)
\(\Rightarrow \int ^{\frac{\pi}{2}}_{\frac{\pi}{3}}(2-\cot ^2x)dx=\left.\begin{matrix} \frac{\pi}{2}\\ \frac{\pi}{3}\end{matrix}\right|(3x+\cot x+c)=\frac{\pi}{2}-\frac{\sqrt{3}}{3}\)
3)
Xét \(\int (2\tan x-3\cot x)^2dx\)
\(=\int (4\tan ^2x+9\cot ^2x-12)dx\)
\(=\int (\frac{4\sin ^2x}{\cos ^2x}+\frac{9\cos ^2x}{\sin ^2x}-12)dx\)
\(=\int (\frac{4(1-\cos ^2x)}{\cos ^2x}+\frac{9(1-\sin ^2x)}{\sin ^2x}-12)dx\)
\(=\int (\frac{4}{\cos ^2x}+\frac{9}{\sin ^2x}-25)dx\)
\(=4\int d(\tan x)-9\int d(\cot x)-25\int dx\)
\(=4\tan x-9\cot x-25x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(2\tan x-3\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(4\tan x-9\cot x-25x+c)=\frac{26\sqrt{3}}{3}-\frac{25\pi}{6}\)
2)
Xét \(\int (\tan x+\cot x)^2dx=\int (\tan ^2x+\cot ^2x+2)dx\)
\(=\int (\frac{\sin ^2x}{\cos^2 x}+\frac{\cos ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1-\cos ^2x}{\cos ^2x}+\frac{1-\sin ^2x}{\sin ^2x}+2)dx\)
\(=\int (\frac{1}{\cos ^2x}+\frac{1}{\sin ^2x})dx\)
\(=\int d(\tan x)-\int d(\cot x)=\tan x-\cot x+c\)
Do đó:
\(\int ^{\frac{\pi}{3}}_{\frac{\pi}{6}}(\tan x+\cot x)^2dx=\left.\begin{matrix} \frac{\pi}{3}\\ \frac{\pi}{6}\end{matrix}\right|(\tan x-\cot x+c)=2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)
\(=2^3+30-\dfrac{3}{2}\)
\(=36,5\)
b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)
\(=0,1^{-1}-2^2-2^{-4}\)
\(=10-4-\dfrac{1}{16}\)
\(=\dfrac{95}{16}\)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)
a.
\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)
b.
\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)
c.
\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)
d.
\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)
e.
\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)
g.
\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)
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g)Ta có f(x) = sin3xcos5x là hàm số lẻ.
Vì f(-x) = sin(-3x)cos(-5x) = -sin3xcos5x = f(-x) nên: