Ta có :\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

=> \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

=> \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

=> \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

=> x + 20 = 0

=> x = -20

Vậy x = -20

\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

\(\Leftrightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

\(\Rightarrow x+20=0\Rightarrow x=-20\)

cái bên vế phải chả hiểu là j ?? Tớ tạm đặt là    \(\alpha\)    nhé      (   \(\alpha\)là 1 hằng số    )

pt <=>   \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=\alpha\)

<=>   \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=\alpha\)

<=>   \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=\alpha\)

<=>   \(\frac{\left(x+20\right).181}{660}=\alpha\)

<=>   \(x=\frac{660\alpha}{181}-20\)

Bạn tự thay    \(\alpha\)là 1 số nào tùy ý cũng sẽ tìm ngay ra x thôi 

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{21}\)và \(5x+y-2z=28\)

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

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

=> 4x + 4 = 3x - 6

=> 4x - 3x = - 6 - 4

=> x = - 10

b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)

\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)

\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0

\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)

Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0

=> x = -1

c) Xem lại đề

x=-1               

x=-1                                 

x = -1

ai tk mk

mk tk lại

hứa luôn

thank nhiều

Công mỗi phân số cho 1 .....................

 mỗi hạng tử ở 2 vế cộng với 1 (có nghĩa là cộng 2 vế với 3 xong chia đều ra 3 hạng tử mỗi hạng tử cộng với 1)

Sau đó sẽ dẫn đến tất cả các hạng tử đều có chung tử số rồi nhóm tử ra ngoài là được 

\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)

Cộng mỗi p/s cho 1,ta đc:

\(\frac{x-6}{7}+1+\frac{x-7}{8}+1+\frac{x-8}{9}+1=\frac{x-9}{10}+1+\frac{x-10}{11}+1+\frac{x-11}{12}+1\)

\(\Leftrightarrow\frac{x-6+7}{7}+\frac{x-7+8}{8}+\frac{x-8+9}{9}=\frac{x-9+10}{10}+\frac{x-10+11}{11}+\frac{x-11+12}{12}\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\left(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\right)=0\)

\(\Leftrightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)

Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\)

=>x+1=0

=>x=-1

\(ĐKXĐ:x\ne3;x\ne5;x\ne4;x\ne6\)

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

\(\Rightarrow\frac{x}{x-3}-\frac{x}{x-5}-\frac{x}{x-4}+\frac{x}{x-6}=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\left(tm\right)\\\frac{1}{x-3}-\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}=0\left(1\right)\end{cases}}\)

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

\(\Rightarrow\frac{2x-9}{\left(x-3\right)\left(x-6\right)}=\frac{2x-9}{\left(x-5\right)\left(x-4\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\left(tm\right)\\\left(x-3\right)\left(x-6\right)=\left(x-5\right)\left(x-4\right)\left(2\right)\end{cases}}\)

\(\left(2\right)\Leftrightarrow x^2-9x+18=x^2-9x+20\)

\(\Leftrightarrow0=2\left(L\right)\)

Vậy pt có 2 nghiệm \(\left\{0;\frac{9}{2}\right\}\)

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

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

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

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

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

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

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

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

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

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

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

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

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

a/ ĐKXĐ: \(x\ne\left\{8;9;10;11\right\}\)

\(\frac{8}{x-8}+1+\frac{11}{x-11}+1=\frac{9}{x-9}+1+\frac{10}{x-10}+1\)

\(\Leftrightarrow\frac{x}{x-8}+\frac{x}{x-11}=\frac{x}{x-9}+\frac{x}{x-10}\)

\(\Leftrightarrow x\left(\frac{1}{x-8}-\frac{1}{x-9}+\frac{1}{x-11}-\frac{1}{x-10}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-9}-\frac{1}{x-8}=\frac{1}{x-11}-\frac{1}{x-10}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\frac{1}{\left(x-9\right)\left(x-8\right)}=\frac{1}{\left(x-11\right)\left(x-10\right)}\)

\(\Leftrightarrow x^2-17x+72=x^2-21x+110\)

\(\Rightarrow x=\frac{19}{2}\)

b/ ĐK: \(x\ne\left\{3;4;5;6\right\}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x-3}-\frac{1}{x-5}=\frac{1}{x-4}-\frac{1}{x-6}\left(2\right)\end{matrix}\right.\)

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

\(\Leftrightarrow x^2-8x+15=x^2-10x+24\)

\(\Rightarrow x=\frac{9}{2}\)

a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x² - 4x + 8 - 2x² - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

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

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}