A=\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x-3\right)\left(x+3\right)\left(x^2+1\right)}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\) ( với \(x^4-8x^2-9=x^4-9x^2+x^2-9=x^2\left(x^2-9\right)+\left(x^2-9\right)=\left(x^2-9\right)\left(x^2+1\right)=\left(x-3\right)\left(x+3\right)\left(x^2+1\right)\)  

A= \(\frac{13-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x+3}-\frac{2}{x-3}=0\) \(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\) \(\Leftrightarrow\left(10x-30\right)\left(x-3\right)+6-2\left(x+3\right)=0\Leftrightarrow-x^2+11x-30=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=5\end{array}\right.\)

Mày nhìn cái chóa j

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

= \(x^8.\frac{1}{10}.\frac{2}{9}.\frac{3}{8}.\frac{4}{7}.\frac{5}{6}.\frac{6}{5}.\frac{7}{4}.\frac{8}{3}.\frac{9}{2}\)

= \(x^8.\frac{1}{10}.\left(\frac{2}{9}.\frac{9}{2}\right).\left(\frac{3}{8}.\frac{8}{3}\right).\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{5}{6}.\frac{6}{5}\right)\)

= \(x^8.\frac{1}{10}.1.1.1.1\)

= \(x^8.\frac{1}{10}\)

Mk ko pik co dung ko nua

=4/10

\(\frac{x^8}{10}\)

\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{x^2+5x+6}-\frac{2}{x-3}=0\)

\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{x^4-8x^2+16-25}-\frac{3\left(x+2\right)}{x^2+2x+3x+6}-\frac{2}{x-3}=0\)

\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^4-8x^2+16\right)-5^2}-\frac{3\left(x+2\right)}{x\left(x+2\right)+3\left(x+2\right)}-\frac{2}{x-3}=0\)

\(\Leftrightarrow\frac{13-x}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^2-4\right)^2-5^2}-\frac{3\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}-\frac{2}{x-3}=0\)

\(\Leftrightarrow\frac{13-x}{x+3}-\frac{3}{x+3}+\frac{6\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-9\right)}-\frac{2}{x-3}=0\)

\(\Leftrightarrow\frac{10-x}{x+3}+\frac{6}{\left(x-3\right)\left(x+3\right)}-\frac{2}{x-3}=0\)

\(\Leftrightarrow\frac{\left(10-x\right)\left(x-3\right)+6-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow10x-30-x^2+3x+6-2x-6=0\)

\(\Leftrightarrow-x^2+11x-30=0\)

\(\Leftrightarrow-x^2+5x+6x-30=0\)

\(\Leftrightarrow-x\left(x-5\right)+6\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(-x+6\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-5=0\\-x+6=0\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=5\\x=6\end{matrix}\right.\)

Vậy x=5 ;x=6

Phương trình này k có nghiệm

(A Trọng ns thế)

ĐK: \(x\ne-3,3,-2\)

Ta có: \(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-8x^2-9}-\frac{3x+6}{x^2+5x+6}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^4-9x^2+x^2-9}-\frac{3x+6}{x^2+3x+2x+6}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6x^2+6}{x^2.\left(x^2-9\right)+\left(x^2-9\right)}-\frac{3x+6}{x.\left(x+3\right)+2.\left(x+3\right)}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6.\left(x^2+1\right)}{\left(x^2+1\right).\left(x^2-9\right)}-\frac{3.\left(x+2\right)}{\left(x+2\right).\left(x+3\right)}-\frac{2}{x-3}=0\)

=>\(\frac{13-x}{x+3}+\frac{6}{x^2-9}-\frac{3}{x+3}-\frac{2}{x-3}=0\)

=>\(\left(\frac{13-x}{x+3}-\frac{3}{x+3}\right)+\left(\frac{6}{x^2-9}-\frac{2}{x-3}\right)=0\)

=>\(\frac{13-x-3}{x+3}+\left[\frac{6}{x^2-9}-\frac{2.\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\right]=0\)

=>\(\frac{10-x}{x+3}+\left[\frac{6}{x^2-9}-\frac{2x+6}{x^2-9}\right]=0\)

=>\(\frac{10-x}{x+3}+\frac{6-2x-6}{x^2-9}=0\)

=>\(\frac{\left(10-x\right).\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}+\frac{-2x}{x^2-9}=0\)

=>\(\frac{13x-x^2-30}{x^2-9}-\frac{2x}{x^2-9}=0\)

=>\(\frac{13x-x^2-30-2x}{x^2-9}=0\)

=>\(\frac{11x-x^2-30}{x^2-9}=0\)

Vì \(x\ne-3,3=>x^2\ne0\)

=>11x-x²-30=0

=>6x-30-x²+5x=0

=>6.(x-5)-x.(x-5)=0

=>(6-x).(x-5)=0

=>6-x=0=>x=6

hoặc x-5=0=>x=5

Vậy tập nghiệm của phương trình S=6; 5

Em ước gì được ên lớp 8 để giúp anh  Hoàng Phúc

Nobita Kun nó lớp 7 đó cha ko làm thì thôi ai bắt vô cmt

4TTTT6 

gì v bn

1/ \(\frac{3\left(x+3\right)}{4}+\frac{1}{2}=\frac{5x+9}{3}-\frac{7x-9}{4}\)

=> \(\frac{9\left(x+3\right)}{12}+\frac{6}{12}=\frac{4\left(5x+9\right)}{12}-\frac{3\left(7x-9\right)}{12}\)

=> \(9\left(x+3\right)+6=4\left(5x+9\right)-3\left(7x-9\right)\)

=> \(9x+27+6=20x+36-21x+27\)

=> \(9x-20x+21x=27-27-6+36\)

=> \(10x=30\)

=> \(x=3\)

Vậy phương trình có tập nghiệm là \(S=\left\{3\right\}\)

2.Ta có : \(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)

=> \(\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{510}{30}\)

=> \(10\left(2x-3\right)-5\left(x-3\right)=6\left(4x+3\right)-510\)

=> \(20x-30-5x+15=24x+18-510\)

=> \(20x-5x-24x=18-510+30-15\)

=> \(-9x=-477\)

=> \(x=53\)

Vậy phương trình có tập nghiệm là \(S=\left\{53\right\}\)

3/ Ta có : \(\frac{5x-1}{6}+\frac{2\left(x+4\right)}{9}=\frac{7x-5}{15}+x-1\)

=> \(\frac{30\left(5x-1\right)}{180}+\frac{40\left(x+4\right)}{180}=\frac{12\left(7x-5\right)}{180}+\frac{180x}{180}-\frac{180}{180}\)

=> \(30\left(5x-1\right)+40\left(x+4\right)=12\left(7x-5\right)+180x-180\)

=> \(150x-30+40x+160=84x-60+180x-180\)

=> \(150x+40x-180x-84x=-60-180-160+30\)

=> \(-74x=-370\)

=> \(x=5\)

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