Ta có: \(4a=3b=2c\Rightarrow\frac{4a}{12}=\frac{3b}{12}=\frac{2c}{12}\Rightarrow\frac{a}{3}=\frac{b}{4}=\frac{c}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{6}=\frac{a+b+c}{3+4+6}=\frac{169}{13}=13\)

\(\frac{a}{3}=13\Rightarrow a=39;\frac{b}{4}=13\Rightarrow b=52;\frac{c}{6}=13\Rightarrow c=78\)

a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\frac{4a-3b}{4a+3b}=\frac{4c-3d}{4c+3d}\Rightarrow\frac{4a-3d}{4c-3d}=\frac{4a+3b}{4c+3d}\)

b) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)

Áp dụng dãy tỉ số bằng nhau ta có :

\(\frac{2a}{3b}=\frac{2c}{2d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)

a) 3a + 4b - 5c - 2a - 3b + 5c

= ( 3a - 2a ) + ( 4b - 3b ) - ( 5c - 5c )

= a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= ( 7a - 3a - 4a ) + ( 3b + 2b + b ) - ( 4c + 2c + 2c ) 

= 6b - 8c

a) 3a + 4b - 5c - 2a - 3b + 5c

= (3a - 2a) + (4b - 3b) - (5c - 5c)

= a + b - 0 = a + b

b) 7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

= (7a - 3a - 4a) + (3b + 2b + b) - ( 4c + 2c + 2c)

= 0 + 6b - 8c = 6b - 8c

a)

3a + 4b - 5c - 2a - 3b + 5c

=( 3a - 2a ) + ( 4b - 3b ) + ( -5c + 5c )

= a + b

b)

7a + 3b - 4c - 3a + 2b - 2c - 4a + b - 2c

=( 7a - 3a - 4a ) + ( 3b + 2b + b ) + ( -4c - 2c - 2c )

= 6b + (-8c)

Ta có: a+b=14 =>a=14-b (1)

Thay (1) vào 4a=3b ta được:

4.(14-b)=3b

56-4b=3b

4b+3b=56

7b=56

b=8=>a=8.3:4=6;c=8.3:2=12

Cậu chuyển 4a=3b=2c => =\(\frac{a}{\frac{1}{4}}\)=\(\frac{b}{\frac{1}{3}}\)=\(\frac{c}{\frac{1}{2}}\)

=> \(\frac{a}{\frac{1}{4}}\)=\(\frac{b}{\frac{1}{3}}\)=\(\frac{c}{\frac{1}{2}}\)=\(\frac{a+b}{\frac{1}{3}+\frac{1}{4}}\)=\(\frac{14}{\frac{7}{12}}\)=24

Vì \(\frac{a}{\frac{1}{4}}\)=24=>a=24*\(\frac{1}{2}\)=12

  Tương tư ...

  Dấu này * là dấu nhân nha

\(\dfrac{a}{1}=\dfrac{b}{2}=\dfrac{c}{3}=\dfrac{4a-3b+2c}{4-6+6}=\dfrac{36}{4}=9\\ \Rightarrow\left\{{}\begin{matrix}a=9\\b=18\\c=27\end{matrix}\right.\\ \dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{16}=\dfrac{x-y+z}{10-15+16}=\dfrac{-49}{11}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{490}{11}\\y=-\dfrac{735}{11}\\z=-\dfrac{784}{11}\end{matrix}\right.\)

a) Có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{4a}{3b}=\frac{4c}{3d}\)

Áp dụng tỉ lệ thức ta có :

\(\frac{4a}{3b}=\frac{4c}{3d}\Rightarrow\)\(\frac{4a}{4c}=\frac{3b}{3d}\Rightarrow\frac{4a+3b}{4c+3d}=\frac{4c-3d}{4c-3d}\)

b) Có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{2a}{3b}=\frac{2c}{3d}\)

Áp dụng tỉ lệ thức ta có "

\(\frac{2a}{3b}=\frac{2c}{3d}\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\Rightarrow\frac{2a-3b}{2c-3d}=\frac{2a3b}{2c+3d}\Rightarrow\frac{2a-3b}{2a+3b}=\frac{2c-3d}{2c+3d}\)

Các câu còn lại bạn làm tương tự

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

\(\left(2a+3b\right)\left(4c-5d\right)=\left(4a-5b\right)\left(2c+3d\right)\)

\(\Leftrightarrow8ac-10ad+12bc-15bd=8ac+12ad-10bc-15bd\)

\(\Leftrightarrow-10ad+12bc=12ad-10bc\)

\(\Leftrightarrow\left(-10ad+12bc\right)+\left(-12bc-12ad\right)=\left(12ad-10bc\right)+\left(-12bc-12ad\right)\)

\(\Leftrightarrow22bc=22ad\)