a)\(\frac{x}{108}=\frac{-7}{9}.\frac{5}{6}\)

\(\frac{x}{108}=\frac{-35}{54}\)

\(\frac{x}{108}=\frac{-70}{108}\)

\(x=-70\)

b) 

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

\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)

\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau

Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

tự làm để nc trí óc dễ

\(3x+\frac{3}{9}+\frac{5}{8}+1=6\)

\(\Rightarrow3x+\frac{47}{24}=6\)

\(\Rightarrow3x=\frac{97}{24}\)

\(\Rightarrow x=\frac{97}{72}\)

\(x+\frac{3}{9}+x+\frac{5}{8}+x+\frac{7}{7}=6\)

\(3x=6-\frac{3}{9}-\frac{7}{7}-\frac{5}{8}\)

\(3x=\frac{97}{24}\)

\(x=\frac{97}{24}\div3\)

\(x=\frac{97}{24}\times\frac{1}{3}\)

\(x=\frac{97}{72}\)

\(x+\frac{3}{9}+x+\frac{5}{8}+x+\frac{7}{7}=6\)

\(x+\frac{1}{3}+x+\frac{5}{8}+x+1=6\)

\(x+\left(\frac{1}{3}+\frac{5}{8}+1\right)=6\)

\(x+\left(\frac{23}{24}+1\right)=6\)

\(x+\frac{47}{24}=6\)

\(x=6-\frac{47}{24}\)

\(x=\frac{97}{24}\)

a)

\(\begin{array}{l}x - \left( { - \frac{7}{9}} \right) =  - \frac{5}{6}\\x + \frac{7}{9} =  - \frac{5}{6}\\x =  - \frac{5}{6} - \frac{7}{9}\\x =  - \frac{{15}}{{18}} - \frac{{14}}{{18}}\\x = \frac{{ - 29}}{{18}}\end{array}\)

Vậy \(x = \frac{{ - 29}}{{18}}\).

b)

\(\begin{array}{l}\frac{{15}}{{ - 4}} - x = 0,3\\x = \frac{{15}}{{ - 4}} - 0,3\\x =  - 3,75 - 0,3\\x =  - 4,05\end{array}\)

Vậy \(x =  - 4,05\).

x=-100

\(\frac{x+5}{95}+\frac{x+6}{94}+\frac{x+7}{93}+\frac{x+8}{92}+\frac{x+9}{91}=-5\)

\(\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)+\left(\frac{x+7}{93}+1\right)+\left(\frac{x+8}{92}+1\right)+\left(\frac{x+9}{91}+1\right)=-5+5=0\)

x=100

k hộ với

\(\frac{1}{2}x+\frac{7}{6}=\frac{9}{5}x-\frac{4}{3}\)

\(\Rightarrow\frac{9}{5}x-\frac{1}{2}x=\frac{7}{6}+\frac{4}{3}\)

\(x.\left(\frac{18}{10}-\frac{5}{10}\right)=\frac{7}{6}+\frac{8}{6}\)

\(\frac{13}{10}.x=\frac{15}{6}\)

\(\Rightarrow x=\frac{15}{6}:\frac{13}{10}\)

\(x=\frac{15}{6}.\frac{10}{13}\)

\(x=\frac{25}{13}\)

Lời giải: Giải phương trình với tập xác định

Tập xác định của phương trình

\(x\in\infty-\infty\)

\(\frac{19x+67}{90}=\frac{15x+83}{56}\Rightarrow\left(19x=67\right)56=90\left(15x+83\right)\)

Kết quả : \(-13\)

kq đúng nhưng mk k biết mấy cái phương trình đó vì mk mới lớp 7

Mashiro Shiina lớp 7 ?????

a) \(x+\left(-7\right)=-20\)

\(\Rightarrow x=-20+7\)

\(\Rightarrow x=-13\)

Vậy \(x=-13\)

b) \(8-x=-12\)

\(\Rightarrow x=8-\left(-12\right)\)

\(\Rightarrow x=20\)

Vậy \(x=20\)

c) \(|x|-7=-6\)

\(\Rightarrow|x|=-6+7\)

\(\Rightarrow|x|=1\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy \(x\in\left\{1;-1\right\}\)

d) \(5^2.2^2-7.|x|=65\)

\(\Rightarrow\left(5.2\right)^2-7.|x|=65\)

\(\Rightarrow10^2-7.|x|=65\)

\(\Rightarrow100-7.|x|=65\)

\(\Rightarrow7.|x|=35\)

\(\Rightarrow|x|=5\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

Vậy \(x\in\left\{5;-5\right\}\)

e) \(37-3.|x|=2^3-4\)

\(\Rightarrow37-3.|x|=8-4\)

\(\Rightarrow37-3.|x|=4\)

\(\Rightarrow3.|x|=33\)

\(\Rightarrow|x|=11\)

\(\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)

Vậy \(x\in\left\{11;-11\right\}\)

f) \(|x|+|-5|=|-37|\)

\(\Rightarrow|x|+5=37\)

\(\Rightarrow|x|=32\)

\(\Rightarrow\orbr{\begin{cases}x=32\\x=-32\end{cases}}\)

Vậy \(x\in\left\{32;-32\right\}\)

g)\(5.|x+9|=40\)

\(\Rightarrow|x+9|=8\)

\(\Rightarrow\orbr{\begin{cases}x+9=8\\x+9=-8\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-17\end{cases}}\)

Vậy \(x\in\left\{-1;-17\right\}\)

h) \(-\frac{5}{6}+\frac{8}{3}+\frac{-29}{6}\le x\le\frac{-1}{2}+2+\frac{5}{2}\)

\(\Rightarrow\frac{-5}{6}+\frac{16}{6}+\frac{-29}{6}\le x\le\frac{-1}{2}+\frac{4}{2}+\frac{5}{2}\)

\(\Rightarrow-3\le x\le4\)

Vậy \(-3\le x\le4\)

câu a

x+(-7)=-20

x=-20-(-7)

x=-13

\(\frac{x+5}{2015}+\frac{x+6}{2014}+\frac{x+7}{2013}+\frac{x+8}{2012}+\frac{x+9}{2011}+5=0\)

\(\Rightarrow1+\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1+\frac{x+8}{2012}+1+\frac{x+9}{2011}=0\)

\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}+\frac{x+2020}{2012}+\frac{x+2020}{2011}=0\)

\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)

\(\Rightarrow x+2020=0\)

\(\Rightarrow x=-2020\)

Study well 

chuyên toán thcs

Thiếu giải thích

Trả lời 

không cần giải thích nha 

\(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}=0\right)\)

\(\Rightarrow\left(x+2020\right)=0:\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)\)

\(\Rightarrow\left(x+2020\right)=0\)

Tôi ngại chuyển vế nên làm tắt có cần giả thích cái j đâu