Lời giải:

PT \(\Leftrightarrow \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+....+\frac{1}{(x+7)(x+8)}=\frac{1}{14}\)

(ĐK: $x\neq -1;-2;...;-8$)

\(\Leftrightarrow \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+....+\frac{(x+8)-(x+7)}{(x+7)(x+8)}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)

\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\Leftrightarrow \frac{7}{x^2+9x+8}=\frac{1}{14}\)

\(\Rightarrow x^2+9x+8=98\Leftrightarrow x^2+9x-90=0\Rightarrow x=6\) hoặc $x=-15$ (đều thỏa mãn)

Vậy........

Bài 1 :

a, Ta có : \(3x-1=2x+4\)

=> \(3x-2x=4+1\)

=> \(x=5\)

Vậy phương trình có tập nghiệm \(S=\left\{5\right\}\)

b, Ta có : \(5x-2=0\)

=> \(5x=2\)

=> \(x=\frac{2}{5}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{2}{5}\right\}\)

c, Ta có : \(7x-4=3x+12\)

=> \(7x-3x=12+4\)

=> \(4x=16\)

=> \(x=4\)

Vậy phương trình có tập nghiệm \(S=\left\{4\right\}\)

d, Ta có : \(\frac{x-1}{2}+\frac{3x+2}{4}=\frac{x-7}{12}\)

=> \(\frac{6\left(x-1\right)}{12}+\frac{3\left(3x+2\right)}{12}=\frac{x-7}{12}\)

=> \(6\left(x-1\right)+3\left(3x+2\right)=x-7\)

=> \(6x-6+9x+6=x-7\)

=> \(6x+9x-x=6-7-6\)

=> \(14x=-7\)

=> \(x=-\frac{1}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{-\frac{1}{2}\right\}\)

Bài 2 :

a, ĐKXĐ : \(\left\{{}\begin{matrix}x^2-2x+1\ne0\\x-1\ne0\end{matrix}\right.\)

=> \(x-1\ne0\)

=> \(x\ne1\)

- Ta có : \(\left(\frac{x+1}{x^2-2x+1}+\frac{1}{x-1}\right):\frac{x}{x-1}-\frac{2}{x-1}\)

= \(\left(\frac{x+1}{\left(x-1\right)^2}+\frac{x-1}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)

= \(\left(\frac{2x}{\left(x-1\right)^2}\right):\frac{x}{x-1}-\frac{2}{x-1}\)

= \(\left(\frac{2x}{\left(x-1\right)^2}\right)\left(\frac{x-1}{x}\right)-\frac{2}{x-1}\)

= \(\frac{x}{x-1}-\frac{2}{x-1}\)

= ​​\(\frac{x-2}{x-1}\)

a/ \(\left|\frac{3x-6}{1-2x}\right|=x-2\) \(\left(x\ne\frac{1}{2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{3x-6}{1-2x}=x-2\\\frac{3x-6}{1-2x}=2-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=\left(x-2\right)\left(1-2x\right)\\3x-6=\left(2-x\right)\left(1-2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=x+4x-2-2x^2\\3x-6=-x-4x+2+2x^2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2x^2+2x+4=0\\2x^2-8x+8=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

KL: .............

b/ Tương tự

a. \(\frac{5x-2}{3}=\frac{5x-3x}{2}\)
\(\Leftrightarrow2.\left(5x-2\right)=3.\left(5x-3x\right) \)
\(\Leftrightarrow10x-4=15x-9x\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy...
b. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\left(1\right)\)
MC = 36.
pt (1) <=>
\(\frac{3\left(10x+3\right)}{36}=\frac{36}{36}+\frac{4\left(6+8x\right)}{36}\)
=> 3.(10x+3) = 36 + 4(6+8x)
<=> 30x+9 = 36+24+32x
<=> -2x = 51
<=> x = \(\frac{-51}{2}\)
Vậy...
c. \(\frac{7x-1}{6}+2=\frac{16-x}{5}\left(2\right)\)
MC = 30.
pt (2) <=>
\(\frac{5\left(7x-1\right)}{30}+\frac{60x}{30}=\frac{6\left(16-x\right)}{30}\)
=> 5(7x-1) + 60x = 6(16-x)
<=> 35x-5 + 60x = 96-6x
<=> 101x = 101
<=> x = 1
Vậy...
d. \(\frac{3x+2}{2}-\frac{3x+1}{6}=5\) (3)
MC = 12.
pt (3)<=>
\(\frac{6\left(3x+2\right)}{12}-\frac{2\left(3x+1\right)}{12}=\frac{60}{12}\)
=> 6(3x+2) - 2(3x+1) = 60
<=> 18x+12 - 6x-2 = 60
<=> 12x = 50
<=> x = \(\frac{25}{6}\)
Vậy...
e. \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\) (4)
MC = 30.
pt (4) <=>
\(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> 6(x+4) - 30x + 120 = 10x - 15(x-2)
<=> 6x+24 - 30x + 120 = 10x - 15x+30
<=> -19x = -114
<=> x = \(\frac{114}{19}=6\)
Vậy...

ai làm ơn giúp tớ đi mà TwT