Ta có:

\(\left\{{}\begin{matrix}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\\...\\\left|x+\frac{1}{110}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\ge0\)

\(\Rightarrow11x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{110}\right|\)

=\(x+\frac{1}{2}+x+\frac{1}{6}+...+x+\frac{1}{110}\)

\(=10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)

Đặt \(A=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\)

\(\Rightarrow A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{11-10}{10.11}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(\Rightarrow A=1-\frac{1}{11}=\frac{10}{11}\)

\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=10x+A=10x+\frac{10}{11}=11x\)

\(\Rightarrow\frac{10}{11}=11x-10x\)

\(\Rightarrow x=\frac{10}{11}\)

Ta có: \(\left|x+\frac{1}{2}\right|\ge0\left|x+\frac{1}{6}\right|\ge0;...;\left|x+\frac{1}{110}\ge0\right|\)

\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{100}\right|\ge0\)

\(\Rightarrow11x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{2}>0;x+\frac{1}{6}>0;...;x+\frac{1}{100}>0\)

\(\Rightarrow\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{6}\right|=x+\frac{1}{6};...;\left|x+\frac{1}{100}\right|=x+\frac{1}{110}\)

\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)=11x\)

\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=11x\)

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow10x+\frac{10}{11}=11x\)

\(\Rightarrow x=\frac{10}{11}\)

vì |x+1/2| ; |x+1/6| ; ............ ; |x+110| lớn hơn hoặc bằng 0=> 11x lớn hớn hoặc bằng 0=> x lớn hớn hoặc bằng 0

=>x+1/2 ; x+1/6 ; ............ ; x+110 lớn hơn hoặc bằng 0

ta có: x+1/2+x+1/6+x+1/12+...+x+1/110=11x

(x+x+...+x)+(1/1.2+1/2.3+1/3.4+...+1/10.11)=11x

10x+(1-1/10)=11x

x= 1/9

à mình bỏ dấu" | " vì khi mà lớn hơn hoặc bằng 1 rồi thfi bỏ ra nó vẫn có giá trị bằng giá trị trị lúc ban đầu

Tìm x, biết:

3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)

2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)

x+110 +2+111 x+112 =x+113 +x+114 

x−1030 +x−1443 +x−595 +x−1488 =0

Trả lời luôn à bạn

Vì \(\left|x+\frac{1}{2}\right|\ge0;\left|x+\frac{1}{6}\right|\ge0;\left|x+\frac{1}{12}\right|\ge0;...;\left|x+\frac{1}{110}\right|\ge0\)

\(\Rightarrow11x\ge0\)

\(\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{110}\right)=11x\)

\(\Rightarrow\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{10.11}\right)=11x\)

\(\Rightarrow\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=11x\)

10 số x

\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=11x\)

\(\Rightarrow1-\frac{1}{11}=11x-10x\)

\(\Rightarrow x=\frac{10}{11}\)

Vậy \(x=\frac{10}{11}\)

ngu như con bò tót, ko biết 1+1=2.

ngu như con bò tót, ko biết 1+1=2.

ngu như con bò tót, ko biết 1+1=2.

M viết j đó

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-8}+\frac{1}{x-8}-\frac{1}{x-20}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{2}{x-20}=-\frac{3}{4}\)

\(\frac{x-20-2x+2}{x^2-21x+20}=\frac{-3}{4}\)

\(-4x-72=-3x^2+63x-60\)

\(-3x^2+63x-60+4x+72=0\)

\(-3x^2+67x+12=0\)

Ko có x t/m

Mình đang cần gắp 

bạn còn

b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

A = ( 1 - 1/2) . ( 1 - 1/3 ) . (1-1/4) ....(1-1/2015) . (1-1/2016)

A= 1/2 . 2/3 . 3/4...2014/2015 . 2015/2016

A = 1 . 2 . 3 . 4 ... 2014 . 2015/ 2 . 3 . 4 ... 2015 . 2016

A = 1/ 2016