\(\frac{1}{x}+\frac{1}{2.x}+\frac{1}{6x}+\frac{1}{12x}+\frac{1}{30x}\)

= \(\frac{1}{x}\left(1+\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{30}\right)\)

= \(\frac{1}{x}\left(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)

= \(\frac{1}{x}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)

=\(\frac{1}{x}\left(1+1-\frac{1}{6}\right)\)

=\(\frac{1}{x}.\frac{11}{6}\)

=\(\frac{11}{6x}\)

\(x=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{5}{6}\)

x-1/2-1/6-1/12=1/30+1/20

x-1/2-1/6-1/12=1/12

x-1/2-1/6=1/12+1/12

x-1/2-1/6=1/6

x-1/2=1/6+1/6

x-1/2=1/3

x=1/3+1/2

x=2/6+3/6

x=5/6

`x - 1/2 - 1/6 - 1/12 - 1/20 = 1/30`

`=> x   = 1/30 + 1/20 + 1/12 + 1/6 + 1/2`

`=> x  = 5/6`

vậy `x = 5/6`

1/2 + 1/6 + 1/12 + 1/20 + 1/30 1/x = 41/42

5/6 + 1/x = 41/42

1/x = 41/42 - 5/6 

1/x = 1/7 

vậy x = 7

x = 7 ở vong 18 đúng không

chuẩn 

x + 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 = 1 

x + 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 = 1 

x + 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 = 1 

x + 1/1 - 1/7 = 1 

x + 6/7 = 1 

x = 1 - 6/7 

x = 1/7 

x + 1/2 + 1/6 + 1/20 + 1/30 + 1/42 = 1

x + 65/84 = 1

x = 1 - 65/84

x = 19/84

x + 1/2 + 1/6 + 1/12 + 1/20 + 1/30 × 1/42 = 1

<=> x + 6/7 = 1

<=> x = 1 - 6/7

<=> x = 1/7

 Vậy x = 1/7

bằng 7 dúng không

\(\frac{1}{x}=\frac{41}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(\frac{1}{x}=\frac{1}{7}=>x=7\)

ĐK : \(x\ne-2.-3;-4;-5;-6\)

\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\Leftrightarrow\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\Leftrightarrow x^2+8x-20=0\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow x=2;x=-10\)( tmđkxđ )

Vậy tập nghiệm phương trình là S = { -10 ; 2 } 

ĐKXĐ \(x\notin\left\{-2;-3;...;-6\right\}\)

Phương trình tương đương với:

\(\dfrac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\dfrac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\\\Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\\ \Leftrightarrow\begin{matrix}x=2\\x=-10\end{matrix}\left(t.m\right)\)

ĐKXĐ: \(x\notin\left\{-2;-3;-4;-5;-6\right\}\)

Ta có: \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{8\left(x+6\right)}{8\left(x+2\right)\left(x+6\right)}-\dfrac{8\left(x+2\right)}{8\left(x+2\right)\left(x+6\right)}=\dfrac{\left(x+2\right)\left(x+6\right)}{8\left(x+2\right)\left(x+6\right)}\)

Suy ra: \(8x+48-8x-16=x^2+8x+12\)

\(\Leftrightarrow x^2+8x+12-32=0\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-10;2}

41/42 - 1/2 - 1/6 - 1/12 - 1/20 -1/30 = 1/x

1/7 = 1/x

x = 1 / 1/7

x = 7

chi lớp 4 há

                          \(\frac{1}{1X2}+\frac{1}{2X3}+\frac{1}{3X4}+\frac{1}{4X5}+\frac{1}{5X6}+\frac{1}{x}\)=    \(\frac{41}{42}\)

                            \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{x}=\frac{41}{42}\)

                             \(1-\frac{1}{6}+\frac{1}{x}=\frac{41}{42}\)

                            \(\frac{5}{6}+\frac{1}{x}=\frac{41}{42}\)     

                                             \(\frac{1}{x}\)     =\(\frac{41}{42}-\frac{5}{6}\) 

                                              \(\frac{1}{x}\)    =   \(\frac{1}{7}\)        

                         VẬy x = 7

                       k cho mik nha

đề lạ thế