Giải : 

Từ đẳng thức\(\frac{a}{a+c}=\frac{b}{b+d}\)

\(\Rightarrow a.\left(b+d\right)=\left(a+c\right).b\)

\(\Rightarrow ab+ad=ab+cb\)

\(\Rightarrow ad=cb\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)

Ta có:

\(\frac{a}{a+c}=\frac{b}{b+d}\Rightarrow\frac{a}{b}=\frac{a+c}{b+d}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{a}{b}=\frac{a+c}{b+d}=\frac{a+c-a}{b+d-b}=\frac{c}{d}\)

=> \(\frac{a}{b}=\frac{c}{d}\).

a) \(-\frac{3}{x}=\frac{15}{7}\)
=> -3.7 = 15x
=> 15x = -21
=>     x = -21:15
=>     x = -1,4
Vậy x = -1,4
b) \(\frac{x+3}{4}=\frac{5}{20}\)
\(\Rightarrow\frac{x+3}{4}=\frac{1}{4}\)
=> x + 3 = 1
=> x       = 1 - 3
=> x       = -2 
Vậy x = -2
d) \(\frac{x-1}{3}=\frac{x+1}{5}\)
=> 5(x - 1) = 3(x + 1)
=> 5x - 5   = 3x + 3
=> 5x - 3x = 5 + 3
=> 2x        = 8
=>   x        = 8:2
=>   x        = 4
Vậy x = 4

\(a,\frac{-3}{x}=\frac{15}{7}\)

=> -21 = 15x

=> \(x=-\frac{21}{15}=-\frac{7}{5}\)

b, 

\(\frac{x+3}{4}=\frac{5}{20}\)

=> \(\frac{5(x+3)}{20}=\frac{5}{20}\)

=> 5\((x+3)\)= 5

=> x + 3 = 1

=> x = -2

\(c,\frac{1,2}{30}=\frac{3x+4}{50}\)

=> \(\frac{\frac{12}{10}}{30}=\frac{3x+4}{50}\)

=> \(\frac{\frac{6}{5}}{30}=\frac{3x+4}{50}\)

=> \(\frac{2}{50}=\frac{3x+4}{50}\)

=> 3x + 4 = 2

=> 3x = -2

=> x = -2/3

\(d,\frac{x-1}{3}=\frac{x+1}{5}\)

=> 5[x - 1] = 3[x + 1]

=> 5x - 5 = 3x + 3 

=> 5x - 5 - 3x = 3

=> 5x - 3x - 5 = 3

=> 2x = 8

=> x = 4

a) Biểu thức trên không có nghĩa khi \(\left(a-1\right)^2=0\)\(\Leftrightarrow a=1\)

b) Khi \(\orbr{\begin{cases}a-2=0\\b+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=2\\b=-5\end{cases}}\)

c) Khi \(a=0\)hoặc \(a=1\)hoặc \(b=0\)

d) Khi \(ab-a^2=0\)\(\Leftrightarrow a\left(b-a\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}a=0\\a=b\end{cases}}\)

\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2019}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2018}{2019}\)

\(A=\frac{1.2.3.....2018}{2.3.4....2019}\)

\(A=\frac{1}{2019}\)

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2019}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2018}{2019}\)

\(A=\frac{1}{2019}\)

Bn ơi tk cho mk nha!!!!!!!!!!!

a) Ta có: 3.4 = 2.6
\(\Rightarrow\frac{3}{2}=\frac{6}{4};\)\(\frac{3}{6}=\frac{2}{4};\)\(\frac{4}{2}=\frac{6}{3};\)\(\frac{4}{6}=\frac{2}{3}\)
b) Ta có: 14.15 = 10.21
\(\Rightarrow\frac{14}{10}=\frac{21}{15};\)\(\frac{14}{21}=\frac{10}{15};\)\(\frac{15}{10}=\frac{21}{14};\)\(\frac{15}{21}=\frac{10}{14}\)
c) Ta có: AB.CD = 2.3
\(\Rightarrow\frac{AB}{2}=\frac{3}{CD};\)\(\frac{AB}{3}=\frac{2}{CD};\)\(\frac{CD}{2}=\frac{3}{AB};\)\(\frac{CD}{3}=\frac{2}{AB}\)
d) Ta có: A.AB = 5.MN
\(\Rightarrow\frac{A}{5}=\frac{MN}{AB};\)\(\frac{A}{MN}=\frac{5}{AB};\)\(\frac{AB}{5}=\frac{MN}{A};\)\(\frac{AB}{MN}=\frac{5}{A}\)

a)3.4=2.6

=\(\frac{3}{2}=\frac{6}{4};\frac{3}{6}=\frac{2}{4};\frac{2}{3}=\frac{4}{6};\frac{6}{3}=\frac{4}{2}\)

b)14.15=10.21

=\(\frac{14}{10}=\frac{21}{15};\frac{14}{21}=\frac{10}{15};\frac{15}{10}=\frac{21}{14};\frac{15}{21}=\frac{10}{14}\)

c)AB.CD=2.3

=\(\frac{AB}{2}=\frac{3}{CD};\frac{AB}{3}=\frac{2}{CD};\frac{CD}{2}=\frac{3}{AB};\frac{CD}{3}=\frac{2}{AB}\)

D)A.AB=5.MN

=\(\frac{A}{5}=\frac{MN}{AB};\frac{A}{MN}=\frac{5}{AB};\frac{AB}{5}=\frac{MN}{A};\frac{AB}{MN}=\frac{5}{A}\)

xy - 2x - 3y = 1

=> (xy - 2x) - 3y + 6 = 1 + 6

=> x(y - 2) - 3(y - 2) = 7

=> (x-3)(y-2) = 7

=> x - 3; y - 2 thuộc Ư(7) = {-1; 1; -7; 7}

ta có bảng : 

\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{6}\) 

\(\frac{x}{4}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\frac{x+y+z}{4+5+6}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\) mà x + y + z = 45

\(\Rightarrow\frac{45}{15}=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow3=\frac{x}{4}=\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\hept{\begin{cases}x=3\cdot4=12\\y=3\cdot6=18\\z=3\cdot5=15\end{cases}}\)

Noob vl

1/9.27ⁿn=3ⁿ

1/9=3ⁿ:27ⁿ

1/9=(1/9)ⁿ

=>n=1