a)

\(\begin{array}{l}\frac{3}{7}.\left( { - \frac{1}{9}} \right) + \frac{3}{7}.\left( { - \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} - \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ - 7}}{9} = \frac{{ - 1}}{3}\end{array}\)                 

b)

\(\begin{array}{l}\left( {\frac{{ - 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ - 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.1 + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 13}}{{13}}\\ = -1\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} - \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ - 2}}{3} + \frac{3}{7} + \frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ - 2}}{3} - \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { - 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)

d)

\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 9}}{15}\\= \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 3}}{5}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{5}{9}.\frac{{ - 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ - 22}}{3} - \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { - 9} \right) =  - 5\end{array}\)

e)

\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ =  - \frac{2}{{97}}\end{array}\)

Theo bài ra , ta có :

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(x\ne3,x\ne-3,x\ne-\frac{7}{2}\)

Quy đồng và khử mẫu phương trình ta đk :

\(13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(13+x-3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow\left(x+3\right)\left(x+10\right)=12x+42\)

\(\Leftrightarrow x^2+13x+30=12x+42\)

\(\Leftrightarrow x^2+13x-12x+30-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

\(\Leftrightarrow\left[\begin{matrix}x-3=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

Kết hợp với ĐKXĐ ta có : x = -4

Vậy \(S=\left\{-4\right\}\)

Chúc bạn học tốt =))

ĐKXĐ: x\(\ne\)3;-7/2;-3

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\Leftrightarrow\frac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\frac{\left(x-3\right)\left(x+3\right)}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\frac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

\(\Leftrightarrow13\left(x+3\right)+\left(x-3\right)\left(x+3\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\\ \Leftrightarrow x^2+x=12\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+4\right)=0\Leftrightarrow\left[\begin{matrix}x-3=0\Rightarrow x=3\\x+4=0\Rightarrow x=-4\end{matrix}\right.\)

Nhận thấy x=3 không thỏa mãn ĐKXĐ nên pt có 1 nghiệm duy nhất là x=-4

Đk: x khác 3; -3 và -7/2

Từ đề suy ra:

13/(x-3)(2x+7) + 1/2x+7 - 6/(x-3)(x+3) = 0

<=> \(\frac{13\left(x+3\right)+\left(x-3\right)\left(x+3\right)-6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}=0\)

<=> 13(x+3) + (x-3)(x+3) + 6(2x+7) = 0

<=> 13x + 39 + x² - 9 + 12x + 42 = 0

<=> x² + 25x + 72 = 0

<=> (x + 25/2)² = 337/4

<=> \(\left[\begin{matrix}x+\frac{25}{2}=\sqrt{\frac{337}{4}}\\x+\frac{25}{2}=-\sqrt{\frac{337}{4}}\end{matrix}\right.\)<=> \(\left[\begin{matrix}x=\frac{-25+\sqrt{337}}{2}\\x=\frac{25-\sqrt{337}}{2}\end{matrix}\right.\)(TM)

Vây ...

13(x+3)+(x+3)(x-3)=6(2x+7)

13x+39+x^2-9-12x-42=0

x^2+x-12=0

x=3 và x=-4

**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

bài này khó ak nha

Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)

<=>1/x+2 -1/x+6 +1/x+6 -1/x+13

<=>1/x+2-1/x+13

=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)

<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)

<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)

<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)

<=> 2x+1=14

<=> 2x=14-1

<=> 2x=13

<=> x=13:2

<=> x=13/2

Vậy x=13/2

Chắc là vầy. Mk cug ko chắc nữa

Vế trái: 4/(x+2).(x+6)+7/(x+6).(x+13)

<=>1/x+2 -1/x+6 +1/x+6 -1/x+13

<=>1/x+2-1/x+13

=> 1/x+2-1/x+13=2x+1/(x+2).(x+16) -3/(x+13).(x+16)

<=>1/x+2 - 1/x+13 + 1/x+13 - 1/x+16=2x+1/(x+2).(x+16)

<=>1/x+2 - 1/x+16=2x+1/(x+2).(x+16)

<=> 14/(x+2).(x+16)= 2x+1/(x+2).(x+16)

<=> 2x+1=14

<=> 2x=14-1

<=> 2x=13

<=> x=13:2

<=> x=13/2

Vậy x=13/2

Chúc bạn học tốt

   \(\frac{13}{\left(2x+7\right)\left(x-3\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\left(1\right)\)

\(ĐKXĐ:x\ne\frac{-7}{2};x\ne\pm3\)

\(MTC:\left(2x+7\right)\left(x-3\right)\left(x+3\right)=\left(2x+7\right)\left(x^2-9\right)\)

\(\left(1\right)\Leftrightarrow\frac{13\left(x+3\right)}{\left(2x+7\right)\left(x^2-9\right)}+\frac{\left(x^2-9\right)}{\left(2x+7\right)\left(x^2-9\right)}=\frac{6\left(2x+7\right)}{\left(2x+7\right)\left(x^2-9\right)}\)

\(\Rightarrow13\left(x+3\right)+\left(x^2-9\right)=6\left(2x+7\right)\)

\(\Leftrightarrow13x+39+x^2-9=12x+42\)

\(\Leftrightarrow13x+x^2+30=12x+42\)

\(\Leftrightarrow x^2+13x-12x+30-42=0\)

\(\Leftrightarrow x^2+x-12\)

\(\Leftrightarrow x^2-3x+4x-12=0\)

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

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

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

Hoặc \(x-3=0\Leftrightarrow x=3\left(L\right)\)

Hoặc \(x+4=0\Leftrightarrow x=-4\left(N\right)\)

Vậy tập nghiệm của phương trình là \(S=\left\{-4\right\}\)

\(\text{ĐKXĐ :}\:x\ne-\frac{7}{2}\:\text{và}\:x\ne\pm3 \). Mẫu chung là \(\left(2x+7\right)\left(x+3\right)\left(x-3\right)\).

Khử mẫu ta được :

\(13\left(x+3\right)+\left(x+3\right)\left(x-3\right)=6\left(2x+7\right)\Leftrightarrow x^2+x-12=0\)

                                                                                               \(\Leftrightarrow x^2+4x-3x-12=0\)

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

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

                                                                                               \(\Leftrightarrow x=-4\:\text{hoặc}\:x=3\)

Trong 2 giá trị tìm được, chỉ có \(x=-4\) là thoả mãn ĐKXĐ. Vậy phương trình có 1 nghiệm duy nhất \(x=-4\).

\(=\frac{6}{7}:\frac{-2}{26}+\frac{6}{7}:\frac{-3}{2}=\frac{-78}{7}+\frac{-4}{7}=\frac{-82}{7}\)

=\(\frac{6}{7}:\left(\frac{3}{26}_{ }-\frac{6}{26}\right)+\frac{6}{7}:\left(\frac{1}{10}-\frac{16}{10}\right)\)

=\(\frac{6}{7}:\frac{-3}{26}+\frac{6}{7}:\frac{-3}{10}\)

=\(\frac{6}{7}.\frac{-26}{3}+\frac{6}{7}.\frac{-10}{3}\)

=\(\frac{6}{7}.\left(\frac{-26}{3}+\frac{-10}{3}\right)\)

=\(\frac{6}{7}.\left(-12\right)\)

=\(\frac{-72}{7}=-10\frac{2}{7}\)

\(=\frac{-52}{7}+\frac{-4}{7}=-8\)