Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)

=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

=> \(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

=> 2x = 12

=> x = 6

Vậy x = 6

\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)                

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)                

\(\frac{6}{2x+1}=\frac{6}{13}\)  

\(\Rightarrow2x+1=13\left(6=6\right)\)         

\(2x=12\)    

\(x=6\)     

1)

a)

\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)

\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)

\(\frac{-20}{5}< x< \frac{-3}{10}\)

\(\frac{-40}{10}< x< \frac{-3}{10}\)

\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)

\(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)

\(\frac{25}{3}< x< \frac{-4}{7}.\frac{1}{1}\)

\(\frac{-25}{3}< x< \frac{-4}{7}\)

\(\frac{-175}{21}< x< \frac{-12}{21}\)

\(\Rightarrow Z\in\left\{-13;-14;-15;-16;...;-174\right\}\)

2)

a)

\(\left(\frac{9}{10}-\frac{15}{16}\right).\left(\frac{5}{12}-\frac{11}{15}-\frac{7}{20}\right)\)

\(=\left(\frac{72}{80}-\frac{75}{80}\right).\left(\frac{25}{60}-\frac{44}{60}-\frac{21}{60}\right)\)

\(=\frac{-3}{80}.\frac{-40}{60}\)

\(=\frac{-1}{-2}.\frac{-1}{-20}\)

\(=\frac{1}{40}\)

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\Rightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)

\(\Rightarrow\frac{21}{18}< \left|x-\frac{12}{18}\right|< \frac{52}{18}\)

còn lại cậu tự tính nha

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\frac{7}{6}< x-\frac{2}{3}< \frac{26}{9}\)

\(\frac{11}{6}< x< \frac{32}{9}\)

a, 2x-1 thuộc ước của 2,rồi giải ra  

b,c tương tự

d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d

mik cảm ơn nhiều nhé mik cx vừa lam ra ạ

\(\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)

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

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

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

\(=\frac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}\)

\(=\frac{x^2-2x+x-2}{\left(x-2\right)\left(x-3\right)}\)

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

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

x^2-5x+6= (x-2)(x-3)

Áp dụng vào biểu thức ta đc

\(=(2x+9)/((x-2)(x-3))-((x+3)(x-3)/((x-2)x-3))+((2x+1)(x-2))/((x-3)(x-2)\)

=(2x+9-x^2+9+2x^2-3x-2)/(x-2)(x-3)

=(16-x+x^2)/(x-2)(x-3)

con lại bạn tự khai triển ra nhé!

Bài 1:

a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)

\(=\frac{-15}{240}-\frac{16}{240}\)

\(=\frac{-31}{240}\)

b, \(=\frac{-10}{12}-\frac{-12}{12}\)

\(=\frac{2}{12}=\frac{1}{6}\)

c, \(=\frac{-30}{6}-\frac{1}{6}\)

\(=\frac{-31}{6}\)

Bài 2:

a, \(x=-\frac{1}{2}-\frac{3}{4}\)

\(x=-\frac{1}{4}\)

b,   \(\frac{1}{2}+x=-\frac{11}{2}\)

\(x=-\frac{11}{2}-\frac{1}{2}\)

\(x=-6\)

Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^

xin lỗi vì chửi hưi quá miệng hahaha

\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)

\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)

\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)

\(=>x=0\)

\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)

\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)

\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)

tu do \(=>x=-7,8;...;0;1;2;3;4\)