sieumau88

$\lim_{n \to \infty}\dfrac{|a_{n+1}|}{|a_{n}|} = \lim_{n \to \infty}\left [ \dfrac{3\left ( n+1 \right )+1}{4\left ( n+1 \right )+2} \cdot \dfrac{3n+1}{4n+2} \right ] = ...... =  \dfrac{9}{16}$
 

Vậy khoảng hội tụ là_ $\left(\dfrac{-16}{9} ; \dfrac{16}{9}\right)$___ $\Leftrightarrow \dfrac{-16}{9} < x < \dfrac{16}{9}$
 
Khi_ $x=\pm \dfrac{16}{9}$ _, chuỗi đã cho có dạng_ $\sum_{n=1}^{+\infty}(-1)^n \cdot \left(\dfrac{3n+1}{4n+2}\right) \cdot \left(\dfrac{16}{9}\right)^n$

 

Ta có__ $\dfrac{|u_{n+1}|}{|u_{n}|} = \left[\dfrac{3(n+1)+1}{4(n+1)+2} \cdot \left(\dfrac{16}{9}\right)^{n+1}\right] : \left[\dfrac{3n+1}{4n+2} \cdot \left(\dfrac{16}{9}\right)^n\right] = ............ = \dfrac{16}{9} > 1$
 

Vậy__ $\sum_{n=1}^{+\infty}(-1)^n \cdot \left(\dfrac{3n+1}{4n+2}\right) \cdot \left(\dfrac{16}{9}\right)^n$__phân kỳ .
 
Kết luận , miền hội tụ của chuỗi đã cho là_ $\left(\dfrac{-16}{9} ; \dfrac{16}{9}\right)$

Câu 3 :

 

$3 .cos 4x - 2 .cos^{2} 3x = 1$

$\Leftrightarrow 3. cos 4x - \left ( 1 + cos 6x \right ) = 1$

$\Leftrightarrow 3. cos 4x - cos 6x - 2 = 0$

 

Đặt__ $t = cos 2x$

 

pt $\Leftrightarrow 3  \left ( 2t^2 - 1 \right ) - \left ( 4t^3 - 3t \right ) - 2 =0$

 

$\rightarrow$ pt bậc 3 theo t $\rightarrow$ giải tìm t  .......

đề bài là gì vậy bạn ?

Câu 1:      $M=x.(y+z)^{2}+y.(z+x)^{2}+z.(x+y)^{2}-4xyz$
$M=[x.(y+z)^{2}-2xyz]+[y.(z+x)^{2}-2xyz]+z.(x+y)^{2}$
$M=x.(y^2+z^2)+y.(x^2+z^2)+z.(x+y)^{2}$
$M=xy.(x+y)+z^2.(x+y)+z.(x+y)^{2}$
$M=(x+y)(xy+z^2+zx+zy)$
$M=(x+y)(x+y)(y+z)$

$A=sin3x+2.sin2x+2.cos2x+2.sinx+4.cosx+3$
$A=(3.sinx-4.sin^3x)+4.sinx.cosx+(2-4sin^2x)+2.sinx+4.cosx+3$
$A=(5.sinx+5)+(-4.sin^3x-4sin^2x)+(4.sinx.cosx+4.cosx)$
$A=5.(sinx+1)-4.sin^2x(sinx+1)+4.cosx.(sinx+1)$
$A=(sinx+1).[5-4.(1-cos^2x)+4.cosx]$
$A=(sinx+1).(4cos^2x+4.cosx+1)$
$A=(sinx+1).(2.cosx+1)^2$

$sin(x+24^{0})+cos(x+144^{0})=cos20^{0}$

 
Đặt       $a=x+84^{0}$

 

pt $\Leftrightarrow sin\left ( a-60^{0} \right )+cos\left ( a+60^{0} \right ) = cos20^{0}$
$\Leftrightarrow sina\cdot \dfrac{1}{2}-cosa\cdot \dfrac{\sqrt{3}}{2}+cosa\cdot \dfrac{1}{2}-sina\cdot \dfrac{\sqrt{3}}{2}= cos20^{0}$
$\Leftrightarrow \dfrac{1}{2}\left ( sina+cosa \right )-\dfrac{\sqrt{3}}{2}\left ( sina+cosa \right )= cos20^{0}$
$\Leftrightarrow sin\left ( a+ 45^{0}\right )\cdot \left (\dfrac{1-\sqrt{3}}{\sqrt{2}} \right )= cos20^{0}$
$\Leftrightarrow sin\left ( x+ 129^{0}\right )= cos20^{0}:\left (\dfrac{1-\sqrt{3}}{\sqrt{2}} \right )$

 

Ko bik quá trình biến đổi có sai sót gì ko nữa ....

Lúc bấm máy tính thấy :     VP  =   -1,815346...  <  -1

Mình chỉ làm đc đến đây thôi .....

Giải phương trình:

$tan^2{2x}.tan^2{3x}.tan^2{5x}=tan^2{2x}-tan^2{3x}-tan^2{5x}$

 

Câu này mình đã giải !! Bạn có thể click đg` link xem tham khảo !!

 

http://diendantoanho...2x-tan23xtan5x/

Tính tổng các nghiệm nằm trong khoảng $(0;2\pi )$ của phương trình:
$(\sqrt{3}-1)sinx+(\sqrt{3}+1)cosx=2\sqrt{2}sin2x$

 

pt $\Leftrightarrow \left (\dfrac{\sqrt{3}-1}{2\sqrt{2}} \right ) \cdot \ sinx \ + \ \left (\dfrac{\sqrt{3}+1}{2\sqrt{2}} \right ) \cdot \ cosx \ = \ sin2x$

$\Leftrightarrow cos\dfrac{5 \pi}{12} \cdot \ sinx \ + \ sin\dfrac{5 \pi}{12} \cdot \ cosx \ = \ sin2x$

$\Leftrightarrow sin\left (x + \dfrac{5 \pi}{12} \right ) = sin2x$

$\Leftrightarrow \left[ \begin{array}{l} 2x = x + \dfrac{5 \pi}{12} + k2 \pi \\ 2x = \pi - x - \dfrac{5 \pi}{12} + k2 \pi \end{array} \right.$ $\left ( k \in \mathbb{Z} \right )$

$\Leftrightarrow \left[ \begin{array}{l} x = \pi \left ( \dfrac{5}{12} + 2k \right ) \\ x = \pi \left ( \dfrac{7}{36} + \dfrac{2k}{3} \right ) \end{array} \right.$ $\left ( k \in \mathbb{Z} \right )$

Vi` điều kiện   $x \in \left ( 0, 2 \pi \right )$   ,  nên phương trình đã cho có nghiệm :

 

$x \in \left \{ \dfrac{5 \pi}{12} \ ; \ \dfrac{5 \pi}{36} \ ; \ \dfrac{31 \pi}{36} \ ; \ \dfrac{55 \pi}{36} \right \}$

Do đó ta có tổng của các nghiệm là

 

$S \ = \ \dfrac{5 \pi}{12} \ + \ \dfrac{5 \pi}{36} \ + \ \dfrac{31 \pi}{36} \ + \ \dfrac{55 \pi}{36} \ = \ \dfrac{53 \pi}{18} $

Tìm các nghiệm x thuộc $\left(0 ; 2 \pi\right)$ của phương trình
$\dfrac{sin3x - sinx}{\sqrt{1 - cos2x}} = sin2x + cos2x$

$\Leftrightarrow \dfrac{2 \ . \ cos2x \ . \ sinx}{\sqrt{2 . sin^2x}} = \sqrt{2} \ . \ cos \left ( 2x - \dfrac{\pi}{4} \right )$

$\Leftrightarrow \dfrac{cos2x  .  sinx}{\left | sinx \right |} = cos \left ( 2x - \dfrac{\pi}{4} \right )$

ĐK : __ $\left\{\begin{matrix}
sinx \neq 0 \\
x \in \left ( 0 ; 2 \pi \right )
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
sinx > 0 \\
x \in \left ( 0 ; 2 \pi \right ) \setminus \left \{ \pi \right \}
\end{matrix}\right.$

 

Dưới đk trên, ta có pt $\Leftrightarrow cos2x = cos \left ( 2x - \dfrac{\pi}{4} \right )$

........v.........v...................

$2\sqrt{2}\cos\left(\dfrac{5\pi}{12}-x \right)\sin x=1$

$\Leftrightarrow 2 . sinx . cos \left(\dfrac{5\pi}{12} - x \right) - \dfrac{1}{\sqrt{2}} = 0$

$\Leftrightarrow sin\dfrac{5\pi}{12} - sin\left(\dfrac{5\pi}{12} - 2x \right) - sin\dfrac{\pi}{4} = 0$

$\Leftrightarrow 2.cos\dfrac{\pi}{3}sin\dfrac{\pi}{12} = sin\left(\dfrac{5\pi}{12} - 2x \right) $

$\Leftrightarrow sin\dfrac{\pi}{12} = sin\left(\dfrac{5\pi}{12} - 2x \right) $

...........v...........v................

$sin^2x +cosx.sin^2x=cos^2x+sinx.cos^2x$

 

$\Leftrightarrow sin^2x - cos^2x + cosx.sin^2x - sinx.cos^2x = 0 $

$\Leftrightarrow \left ( sinx - cosx\right ).\left ( sinx + cosx \right ) + sinx.cosx.\left ( sinx - cosx \right ) = 0$

$\Leftrightarrow \left[ \begin{array}{l} sinx-cosx=0\\ sinx + cosx + sinx.cosx = 0 \end{array} \right.$

$\Leftrightarrow \left[ \begin{array}{l} \sqrt{2}  . sin \left(x - \dfrac{\pi}{4} \right) = 0 \\ t + \dfrac{t^2-1}{2} = 0 \end{array} \right.$ __(Đặt_ $t=sinx + cosx$ __nên ta có__ $sinx.cosx = \dfrac{t^2-1}{2}$

........v......v..................

Ta có___ $ tan5x = \dfrac{tan2x + tan3x}{1-tan2x.tan3x}$

Do đó__ $ tan5x \ . \ \left ( 1 - tan2x.tan3x \right ) = tan2x + tan3x$
________________________________

$\boxed{tan^{2}2x \ . \ tan^{2}3x \ . \ tan5x = tan^{2}2x-tan^{2}3x+tan5x}$

ĐKXĐ : $\left\{\begin{matrix}
tan2x.tan3x \neq 1 \\
cos2x \neq 0 \\
cos3x \neq 0 \\
cos5x \neq 0
\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix}
cosx \neq 0 \\
cos2x \neq 0 \\
cos3x \neq 0 \\
cos5x \neq 0
\end{matrix}\right.$

pt $\Leftrightarrow tan^{2}3x-tan^{2}2x = tan5x \ . \ \left ( 1 - tan^{2}2x \ . \ tan^{2}3x \right )$

$\Leftrightarrow tan^{2}3x-tan^{2}2x = tan5x \ . \ \left ( 1 - tan2x.tan3x \right ) \ . \ \left ( 1+tan2x.tan3x \right )$

$\Leftrightarrow \left (tan3x+tan2x \right ) \ . \ \left (tan3x-tan2x \right ) = \left (tan2x+tan3x \right ) \ . \ \left ( 1+tan2x.tan3x \right )$
 
$\Leftrightarrow \left[ \begin{array}{l} tan2x+tan3x=0\\ tan3x-tan2x = 1+tan2x.tan3x \\ \end{array} \right.$
 
$\Leftrightarrow \left[ \begin{array}{l} tan3x= - tan2x \\ \dfrac{tan3x-tan2x}{1+tan2x.tan3x} = 1 \\ \end{array} \right.$

(Vi` :__ $1+tan2x.tan3x = \dfrac{cos2x.cos3x+sin2x.sin3x}{cos2x.cos3x}= \dfrac{cos5x}{cos2x.cos3x} \neq 0 $ )

 
$\Leftrightarrow \left[ \begin{array}{l} tan3x=tan \left(-2x \right) \\ tanx = tan \dfrac{\pi}{4} \\ \end{array} \right.$
 
.........v.........v...................

$4\cos^4x-cos2x-\frac{1}{2}cos 4x+cos \frac{3x}{4}=\frac{7}{2}$

$\Leftrightarrow \left ( 2\cos^2x \right )^2 - cos2x - \dfrac{cos 4x}{2} + cos \dfrac{3x}{4} = \dfrac{7}{2}$

$\Leftrightarrow \left ( 1+cos2x \right )^2 - cos2x - \dfrac{2cos^2 2x -1}{2} + cos \dfrac{3x}{4} = \dfrac{7}{2}$

$\Leftrightarrow 1 + 2.cos2x + cos^2 2x - cos2x - cos^2 2x + \dfrac{1}{2} + cos \dfrac{3x}{4} = \dfrac{7}{2}$

$\Leftrightarrow cos2x + cos \dfrac{3x}{4} = 2$ ___(**)

Ta có__ $\left\{\begin{matrix}
cos2x \leq 1\\
cos \dfrac{3x}{4} \leq 1
\end{matrix}\right.$ _,_ $\forall x \in \mathbb{R}$
Do đó__ $ cos2x + cos \dfrac{3x}{4} \leq 2$ _,_ $\forall x \in \mathbb{R}$

Dấu "=" của pt (**) xảy ra khi và chỉ khi :

$\Leftrightarrow \left\{\begin{matrix}
cos2x = 1\\
cos \dfrac{3x}{4} = 1
\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x = k \pi\\
x = \dfrac{8k' \pi}{3}
\end{matrix}\right.$ _,_ $\left ( k , k' \in \mathbb{R} \right )$

Giao 2 họ nghiệm trên, ta có nghiệm của pt đã cho là _ $x = 2m \pi$ _,_ $\left ( m \in \mathbb{R} \right )$

$\lim_{n\rightarrow \infty} \left ( \dfrac{1}{1!} +\dfrac{1}{2!}+...+\dfrac{1}{n!}\right) $

 
Ta có___ $ e = \sum_{n=0}^{\infty } \dfrac{1}{n!} = \dfrac{1}{0!} + \sum_{n=1}^{\infty } \dfrac{1}{n!} = 1 + \sum_{n=1}^{\infty } \dfrac{1}{n!}$
Do đó__ $\sum_{n=1}^{\infty } \dfrac{1}{n!} = e - 1 $
Vậy__ $\lim_{n\rightarrow \infty} \left ( \dfrac{1}{1!} +\dfrac{1}{2!}+...+\dfrac{1}{n!}\right ) = \lim_{n\rightarrow \infty} \left ( \sum_{n=1}^{\infty } \dfrac{1}{n!} \right ) = \lim_{n\rightarrow \infty} \left ( e-1 \right ) = e - 1$

$\dfrac{sin^{4}2x+cos^{4}2x}{tan\left ( \dfrac{\pi}{4}-x \right )tan\left ( \dfrac{\pi}{4}+x \right )}=cos^{4}4x$

Xét mẫu số :
$tan\left ( \frac{\pi}{4}-x \right )tan\left ( \frac{\pi}{4}+x \right ) =\dfrac{sin\left ( \frac{\pi}{4}-x \right ) sin\left ( \frac{\pi}{4}+x \right ) }{cos\left ( \frac{\pi}{4}-x \right ) cos\left ( \frac{\pi}{4}+x \right ) }=\dfrac{\frac{1}{2}\left ( cos2x-cos\frac{\pi}{2} \right )}{\frac{1}{2}\left ( cos2x+cos\frac{\pi}{2} \right ) } = \frac{cos2x}{cos2x}$

ĐK: $cos2x \neq 0 $

pt $\Leftrightarrow 1-2sin^{2}2x.cos^{2}2x=cos^{4}4x$

$\Leftrightarrow 1- \dfrac{sin^{2}4x}{2}=cos^{4}4x$

$\Leftrightarrow 2- \left(1-cos^{2}4x \right) = 2.cos^{4}4x$

$\Leftrightarrow 2.cos^{4}4x - cos^{2}4x - 1 = 0$

$\Leftrightarrow \left[ \begin{array}{l} cos^2{4x}=1\\cos^2{4x}= \dfrac{-1}{2} < 0\rightarrow ptvn \end{array} \right.$

$\Leftrightarrow sin^2{4x}=0$

$\Leftrightarrow sin4x=0$

$\Leftrightarrow 2.sin2x.cos2x=0$

$\Leftrightarrow \left[ \begin{array}{l} sin2x=0\\cos2x=0 \rightarrow (Loai)\end{array} \right.$

$\Leftrightarrow sin2x=0 \Leftrightarrow x=\dfrac{k \pi}{2}$ , với $k\in \mathbb{Z}$

 

_______________

 

 

Giải chi tiết giúp bạn !