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

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

\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{8}\)

\(\Leftrightarrow\frac{x-5-x+1}{\left(x-1\right)\left(x-5\right)}=\frac{1}{8}\)

\(\Leftrightarrow-4.8=x^2-6x+5\)

\(\Leftrightarrow x^2-6x+37=0\)

bo tay

kb với tôi đi

\(1.\frac{7x-3}{x-1}=\frac{2}{3}\)   ( \(x\ne1\))

\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)

\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)

\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)

\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-5\)

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

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

\(\Rightarrow\left(1-x\right)\left(3x-1\right)+3\left(x+1\right)\left(3x-1\right)=\left(2x+3\right)\left(x+1\right)\)

\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)

\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)

\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)

\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)

\(\Leftrightarrow4x^2+5x-7=0\)

\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)

\(\left(2x+\frac{5}{4}\right)^2>0\)

\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)

=> PT vô nghiệm 

\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)

\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=\frac{-7}{23}\)

\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)

\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow-6x=-16\)

\(\Leftrightarrow x=\frac{16}{6}\)

\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)

\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)

\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)

\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)

\(\Leftrightarrow x^4+x^3-4x-8=0\)

\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)

Đến đấy mk tắc r xl bạn nhé 

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

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

ĐK:\(x\ne-2;-3;-4;-5\)

MTC:\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right).6\)

Quy đồng khử mẫu:

ĐKXĐ : Tự tìm nha : )

Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)

=> \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)

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

=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)

=> \(\frac{x+8}{\left(x+1\right)\left(x+8\right)}-\frac{x+1}{\left(x+8\right)\left(x+1\right)}=\frac{1}{14}\)

=> \(14\left(x+8-x-1\right)=\left(x+1\right)\left(x+8\right)\)

=> \(x^2+x+8x+8=98\)

=> \(x^2+9x-90=0\)

=> \(\left(x+15\right)\left(x-6\right)=0\)

=> \(\left[{}\begin{matrix}x+15=0\\x-6=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-15\\x=6\end{matrix}\right.\) ( TM )

Vậy phương trình trên có nghiệm là \(S=\left\{6,-15\right\}\)

\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)

\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)

\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)

\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)

\(< =>3072-107x=\frac{38x-684}{5}\)

\(< =>\left(3072-107x\right)5=38x-684\)

\(< =>15360-535x-38x-684=0\)

\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)

nghệm xấu thế 

\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)

\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)

\(< =>993-33x-11x-415=0\)

\(< =>578=44x< =>x=\frac{289}{22}\)

Bài 1: 

b) Phương trình đã cho tương đương với phương trình:

\(\frac{8\left(x+22\right)-55\left(7x+149\right)-6\left(x+12\right)}{45}=\frac{9\left(x+35\right)+2\left(x+50\right)}{45}\)

\(\Leftrightarrow44x=-1056\)

\(\Leftrightarrow x=-24\)

Vậy x=-24 là nghiệm của phương trình

c) Phương trình đã cho tương đương với phương trình:

\(\frac{3x+6}{70}-\frac{x+4}{24}=\frac{32x+19}{60}+\frac{2}{3}\)

\(\Leftrightarrow12\left(3x+6\right)-35\left(x+4\right)=14\left(32x+19\right)+560\)

\(\Leftrightarrow-447x=894\)

\(\Leftrightarrow x=-2\)

Vậy x=-2 là nghiệm của phương trình