ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)

\(\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}=\frac{1}{x+1}+\frac{1}{x+3}+\frac{1}{x+4}+\frac{1}{x+6}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{x+7}+\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+1}+\frac{1}{x+6}+\frac{1}{x+3}+\frac{1}{x+4}\)

\(\Rightarrow\frac{x+7+x}{x\left(x+7\right)}+\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+6+x+1}{\left(x+1\right)\left(x+6\right)}+\frac{x+4+x+3}{\left(x+3\right)\left(x+4\right)}\)

\(\Rightarrow\frac{2x+7}{x^2+7x}+\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+6}+\frac{2x+7}{x^2+7x+12}\)

\(\Rightarrow\left(2x+7\right)\left(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\right)=0\)

mà \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\ne0\)

=> 2x + 7 = 0 => x = -7/2 

                                                                              Vậy x = -7/2

Mày nhìn cái chóa j

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

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

\(\Leftrightarrow\left(x-1\right)\frac{17}{12}=1\)

\(\Leftrightarrow x-1=\frac{12}{17}\)

\(\Leftrightarrow x=\frac{29}{17}\)

Vậy phương trình có tập nghiệm là \(S=\left\{\frac{29}{17}\right\}\).

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

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

\(\Leftrightarrow\frac{6\left(x-1\right)}{12}+\frac{3\left(x-1\right)}{12}=\frac{12}{12}-\frac{4\left(2x-2\right)}{12}\)

\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-4\left(2x-2\right)\)

\(\Leftrightarrow6x-6+3x-3=12-8x+8\)

\(\Leftrightarrow6x+3x+8x=12+8+6+3\)

\(\Leftrightarrow17x=29\)

\(\Leftrightarrow x=\frac{29}{17}\)

Vậy phương trình có nghiệm x=29/17

Đặt u = x - 1

Phương trình trở thành \(\frac{u}{2}+\frac{u}{4}=1-\frac{2u}{3}\)

\(\Leftrightarrow\frac{3u}{4}=\frac{3-2u}{3}\)

\(\Leftrightarrow9u=12-8u\Leftrightarrow17u=12\)

\(\Leftrightarrow u=\frac{12}{17}\Rightarrow x=\frac{12}{17}+1=\frac{29}{17}\)

Vậy phương trình có 1 nghiệm là \(\frac{29}{17}\)

x=-0,44

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

\(\Rightarrow\frac{12x}{12}+\frac{6x+6}{12}+\frac{4x+8}{12}+\frac{3x+9}{12}=\frac{12}{12}\)

\(\Rightarrow25x+23=12\)

\(\Rightarrow x=\frac{-11}{25}\)

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

\(\Rightarrow\frac{12x+6\left(x+1\right)+4\left(x+2\right)+3\left(x+3\right)}{12}=1\)

\(\Rightarrow\frac{12x+6x+6+4x+8+3x+9}{12}=1\)

\(\Rightarrow\frac{25x+23}{12}=1\Rightarrow25x+23=12\Rightarrow x=-\frac{11}{25}\)

Vậy \(x=-\frac{11}{25}\)

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

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

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

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

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

\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)

\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)

\(\Leftrightarrow23x=69\)

\(\Leftrightarrow x=3\)

Vậy nghiệm của pt x=3