(x-5).3/10=2000+5

(x-5).3/10=2005

x-5=2005:3/10

x-5=2005.10/3

x-5=20050/3

x=20050/3 +5

x=20065/3

\(\left(x-5\right)\)\(.\)\(\frac{30}{100}\)\(=\)\(20\times100+5\)

\(\left(x-5\right)\)\(.\)\(\frac{3}{10}\)\(=\)\(2005\)

\(\left(x-5\right)\)\(=\)\(2005\)\(\div\)\(\frac{3}{10}\)

\(\left(x-5\right)\)\(=\)\(2005\)\(.\)\(\frac{10}{3}\)

\(\left(x-5\right)\)\(=\)\(\frac{20050}{3}\)

\(x=\frac{20055}{3}\)

Hok tốt

dễ mà, dựa vào tính chất phân phối của phép nhân đối với phép trừ ( Vế 1)

Tách ra : \(\frac{20}{100}.\frac{x}{1}+5\)

Còn lại như thế nào thj` bn tự tính nhé!!!

\(x=10\)

Mình ko hiểu

a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\) 

\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)

\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)

\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)

Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)

Nên x + 1 = 0

=> x = -1

còn b vs c thì sao ạ

b) Sai đề à bạn đề \(\frac{x+2}{42}+\frac{x+4}{22}=\frac{x+5}{23}+\frac{x+3}{43}\)  hả đề này mk làm đc 

b, 30(x+20) - 6(x-5) - 24 - x =100

30x+600-6x+30-24-x=100

30x-6x+x=100-600-30-24

25x=-554

x=\(-\frac{554}{25}\)

cho li-ke nha

a, -12 . (x-5)+ 7 . ( 3-x ) =15

-12x+60+21-7x=15

-12x7x=15-60-21

-21x=-66

x=  -544/25

tick nha 

\(\frac{5}{14}\frac{5}{14}\) LÀ NHÂN À

Mình lỡ đánh nhầm 2 lần \(\frac{5}{14}\)nha :)) chỉ 1 lần thôi

\(\frac{x-199}{6}+\frac{x-199}{12}+\frac{x-199}{20}+\frac{x-199}{30}+\frac{x-199}{42}=\frac{5}{14}\)

\(\left(x-199\right)\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)=\frac{5}{14}\)

\(\left(x-199\right)\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=\frac{5}{14}\)

\(\left(x-199\right)\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{5}{14}\right)\)

\(\left(x-199\right)\left(\frac{1}{2}-\frac{1}{7}\right)=\frac{5}{14}\)

\(\left(x-199\right).\frac{5}{14}=\frac{5}{14}\)

\(\Leftrightarrow x-199=1\)

\(\Leftrightarrow x=200\)

\(\frac{x+10}{90}+\frac{x+20}{80}+\frac{x+30}{70}+\frac{x+40}{60}+\frac{x+50}{50}=-5\)

<=> \(\frac{x+10}{90}+1+\frac{x+20}{80}+1+\frac{x+30}{70}+1+\frac{x+40}{60}+1+\frac{x+50}{50}+1=0\)

<=> \(\frac{x+100}{90}+\frac{x+100}{80}+\frac{x+100}{70}+\frac{x+100}{60}+\frac{x+100}{50}=0\)

<=> \(\left(x+100\right)\left(\frac{1}{90}+\frac{1}{80}+\frac{1}{70}+\frac{1}{60}+\frac{1}{50}\right)=0\)

<=> x + 100 = 0 

<=> x = -100

Vậy x = -100