từ đề bài: ta có:  1 -  x/17 - 2/17 + x/15 - 4/15 + x/13 - 6/13   =   x/11 - 8/11 + x/9 - 10/9 + x/7 - 12/7

    Tới đây thì tự giải nha

Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)

\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(A=\frac{11}{3}\)(1)

+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)

\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)

=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)

=> \(\frac{11}{3}-x=1\)

=> \(x=\frac{11}{3}-1=\frac{8}{3}\)

Vậy x = 8/3

(x+2)/17+(x+4)/15+(x+6)/13=(x+8)/11+(x+10)/9+(x+12)/7

=>(x+2+17)/17+(x+4+15)/15+(x+6+13)/13=(x+8+11)/11+(x+10+9)/9+(x+12+7)/7

=>(x+19)/17+(x+19)/15+(x+19)/13=(x+19)/11+(x+19)/9+(x+19)/7

=>(x+19)/17+(x+19)/15+(x+19)/13-(x+19)/11-(x+19)/9-(x+19)/7=0

=>(x+19)*(1/17+1/15+1/13-1/11-1/9-1/7)=0

=>x+19=0

=>x=19

áp dụng tc tỉ lệ thức ta có :

\(\Leftrightarrow\frac{671x+2804}{3315}=\frac{239x+2462}{693}\Rightarrow\left(671x+2804\right)693=3315\left(239x+2462\right)\)

=>(671x+2804)693=693(671x+2804) (VT)

<=>693(671x+2804)=3315(239x+2462)

=>465003x+1943172=792285x+8161530

=>-327282x=621835

=>x=621835:(-327282)

=>x=-19

Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.

Phương trình 4:
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Rightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-90-10}{10}+\frac{x-76-24}{12}+\frac{x-58-42}{14}+\frac{x-36-64}{16}+\frac{x-15-85}{17}=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

vế trái có bị sai ko em. chị nghĩ câu hỏi nên là: x+2/17 + x+4/15 + x+6/13=x+8/11 + x+10/9 + x+12/7

Chuyển 2 phân số ở vế kia sang rồi đặt (x+2) ra ngoài....Từ đó có (x+2)=0=>x=-2

Không hiểu phải không???

-2 để vế trên =0 là xong hết mak!!! =)))))) Bà con ủng hộ vài l-i-k-e nhá

\(3x.\left(x-\frac{2}{3}\right)=0\)

\(\Leftrightarrow3x=0\)hoặc \(x-\frac{2}{3}=0\)

\(3x=0\Rightarrow x=0\)

\(x-\frac{2}{3}=0\Rightarrow x=0+\frac{2}{3}=\frac{2}{3}\)

Vậy.. 

\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\\ \Leftrightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=10-10\\ \Leftrightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\\ \Leftrightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)\Leftrightarrow x-100=0\left(Vi\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\right)\Leftrightarrow x=100\)