a) \(x-\frac{5}{7}=\frac{1}{9}\Rightarrow x=\frac{1}{9}+\frac{5}{7}\Rightarrow x=\frac{52}{63}\)

b) \(\frac{-3}{7}-x=\frac{4}{5}+\frac{-2}{3}\Rightarrow\frac{-3}{7}-x=\frac{2}{15}\Rightarrow x=\frac{-3}{7}-\frac{2}{15}\Rightarrow x=\frac{-59}{105}\)

c) \(x-\frac{1}{5}=\frac{2}{7}.\frac{-11}{5}\Rightarrow x-\frac{1}{5}=\frac{-22}{35}\Rightarrow x=\frac{-22}{35}+\frac{1}{5}\Rightarrow x=\frac{-3}{7}\)

d) \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Rightarrow\frac{x}{182}=\frac{-15}{91}\Rightarrow x=\frac{\left(-15\right).182}{91}\Rightarrow x=-30\)

\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

\(b)3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)

\(\Leftrightarrow101x-303=0\)

\(\Leftrightarrow101\left(x-3\right)=0\)

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

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)

\(\Leftrightarrow x=-2020\)

khó lắm

bây h thì bạn giải đc chưa

Cảm ơn bạn rất nhiều mình đã hiểu rồi 

Chúc bạn học tốt nhé

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

       -4                 <x<\(\frac{-3}{10}\)

\(\frac{-40}{10}\)<      x   <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}

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

\(\frac{-875}{189}< x< \frac{108}{189}\)

=> x  E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}

Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.

Phương trình 4:
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Rightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-90-10}{10}+\frac{x-76-24}{12}+\frac{x-58-42}{14}+\frac{x-36-64}{16}+\frac{x-15-85}{17}=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Do \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.

g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)

\(\Leftrightarrow x+100=0\)

\(\Leftrightarrow x=0-100\)

\(\Leftrightarrow x=-100.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)

h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)

\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)

\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)

\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)

\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)

\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)

Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)

\(\Leftrightarrow x-89=0\)

\(\Leftrightarrow x=0+89\)

\(\Leftrightarrow x=89.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)

Chúc bạn học tốt!

Câu g) bạn cộng 1 vào mỗi hạng tử của 2 vế

Câu h) bạn trừ một vào mỗi hạng tử ở hai vế

Quy đồng mẫu thì được tử giống nhau sau đó đặt nhân tử chung là xong