Lời giải:
\(\int \frac{x-1}{\sqrt{x^2-2x+5}}dx=\frac{1}{2}\int \frac{2x-2}{\sqrt{x^2-2x+5}}dx=\frac{1}{2}\int \frac{d(x^2-2x+5)}{\sqrt{x^2-2x+5}}\\ =\frac{1}{2}\int (x^2-2x+5)^{\frac{-1}{2}}d(x^2-2x+5)\\ =\frac{1}{2}.\frac{(x^2-2x+5)^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C=\sqrt{x^2-2x+5}+C\)

Đặt \(t=\sin\left(x\right)\Rightarrow dt=\cos\left(x\right)\)

\(\Rightarrow\int e^{3\sin x}.\cos xdx=\int e^{3t}dt=\dfrac{1}{3}.e^{3t}+C=\dfrac{1}{3}.e^{3\sin x}+C\)

Đặt \(t=lnx\)

=>\(dt=\dfrac{1}{x}dx\)

\(\int\dfrac{1}{xlnx}dx=\int\dfrac{1}{t}dt=ln\left|t\right|=ln\left|lnx\right|\)

=ln(lnx)

\(\int\dfrac{x}{x^2+1}dx\)

Đặt t=x²+1

=>\(dt=2xdx\)

\(\int\dfrac{x}{x^2+1}dx=\dfrac{1}{2}\cdot\int\dfrac{1}{t}dt=\dfrac{1}{2}\cdot\left|t\right|+C=\dfrac{1}{2}\cdot ln\left|x^2+1\right|+C\)

\(=\dfrac{1}{2}\cdot ln\left(x^2+1\right)+C\)

a, A''Có đúng 2 nữ''

\(C^2_3.C_{56}^2\)

\(P\left(A\right)=\dfrac{C_3^2.C_{56}^2}{C_{59}^4}\)

b, B''Có ít nhất 2 nam''

TH1 : Có 2 nam \(C_{56}^2.C_3^2\)

TH2 : Có 3 nam \(C_{56}^3.C_3^1\)

TH3 : Có 4 nam \(C^4_{56}\)

\(\Rightarrow C_{56}^2.C_3^2+C_{56}^3.C_3^1+C_{56}^4\)

\(P\left(B\right)=\dfrac{C_{56}^2.C_3^2+C_{56}^3.C_3^1+C_{56}^4}{C_{59}^4}\)

c, C''Có nhiều nhất 2 nam''

TH1 : Có 1 nam \(C_{56}^1.C_3^3\)

TH2 : Có 2 nam \(C_{56}^2.C_3^2\)

\(\Rightarrow C_{56}^2.C_3^3+C_{56}^2.C_3^2\)

\(P\left(C\right)=\dfrac{C_{56}^2.C_3^3+C^2_{56}.C_3^2}{C_{59}^4}\)

tk

chọn B

?

khong bt

Lấy điểm A bất kì nằm trên đường tròn đáy.

Khi đó góc tạo bởi đường sinh và mặt phẳng đáy chính là \(\widehat{SAO}=45^o\)

Do đó \(h=r=\dfrac{a}{\sqrt{2}}\)

\(\Rightarrow S_{xq}=\pi rl=\pi.\dfrac{a}{\sqrt{2}}.a=\dfrac{\pi a^2}{\sqrt{2}}\)

\(S_{tp}=S_{xq}+\pi r^2=\dfrac{\pi a^2}{\sqrt{2}}+\pi\left(\dfrac{a}{\sqrt{2}}\right)^2=\dfrac{\pi a^2\sqrt{2}+\pi a^2}{2}\) 

`\int [x^2-x+4]/[x^3-3x+2x] dx`

`=\int [x^2-x+4]/[x(x-1)(x-2)] dx`

`=\int (2/x-4/[x-1]+3/[x-2]) dx`

`=2ln|x|-4ln|x-1|+3ln|x-2|+C`.

9: \(\int\dfrac{x^2-x+4}{x^3-3x^2+2x}dx\)

\(=\int\dfrac{\left(x-1\right)\cdot x+4}{x\left(x-1\right)\left(x-2\right)}dx=\int\left(\dfrac{1}{x-2}+\dfrac{4}{x\left(x-1\right)\left(x-2\right)}\right)dx\)

\(=\dfrac{1}{\left|x-2\right|}+\int\dfrac{4}{x\left(x-1\right)\left(x-2\right)}dx\)

Đặt \(\dfrac{4}{x\left(x-1\right)\left(x-2\right)}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x-2}\)

=>\(\dfrac{4}{x\left(x-1\right)\left(x-2\right)}=\dfrac{A\left(x^2-3x+2\right)+B\left(x^2-2x\right)+C\left(x^2-x\right)}{x\left(x-1\right)\left(x-2\right)}\)

=>\(x^2\left(A+B+C\right)+x\left(-3A-2B-C\right)+2A=4\)

=>\(\left\{{}\begin{matrix}A+B+C=0\\-3A-2B-C=0\\2A=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B+C=-A=-2\\3A+2B+C=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}A=2\\B+C=-2\\2B+C=-3A=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=-4\\C=2\end{matrix}\right.\)

=>\(\dfrac{4}{x\left(x-1\right)\left(x-2\right)}=\dfrac{2}{x}+\dfrac{-4}{x-1}+\dfrac{2}{x-2}\)

=>\(\int\dfrac{4}{x\left(x-1\right)\left(x-2\right)}dx=\dfrac{2}{\left|x\right|}+\dfrac{-4}{\left|x-1\right|}+\dfrac{2}{\left|x-2\right|}\)

=>\(\int\dfrac{x^2-x+4}{x^3-3x^2+2x}dx=\dfrac{1}{\left|x-2\right|}+\dfrac{2}{\left|x\right|}+\dfrac{-4}{\left|x-1\right|}+\dfrac{2}{\left|x-2\right|}=\dfrac{3}{\left|x-2\right|}+\dfrac{2}{\left|x\right|}-\dfrac{4}{\left|x-1\right|}\)