b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15) 

Ta có 

lộn

Ta có \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}hay\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có 

 \(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

vậy \(a=12.\dfrac{3}{2}=18\)

\(b=12.\dfrac{4}{3}=16\)

\(c=12.\dfrac{5}{4}=15\)

vậy a=18 ;b=16 ;c=15

\(a,Tacó:\\ \dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=\dfrac{a^3}{2^3}=\dfrac{a\cdot a\cdot a}{2\cdot2\cdot2}=\dfrac{a\cdot b\cdot c}{2\cdot3\cdot5}=\dfrac{810}{30}=27\\ \Rightarrow\left\{{}\begin{matrix}a=27\cdot2=54\\b=27\cdot3=81\\c=27\cdot5=135\end{matrix}\right.\\ Vậy...\)

Các câu khác cx cùng dạng tương tự bn tự làm nha!

a, \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}\) và a . b . c = 810

Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{5}=k\)

=> \(\left\{{}\begin{matrix}a=2k\\b=3k\\c=5k\end{matrix}\right.\)

Mà a . b . c = 810

=> 2k . 3k . 5k = 810

=> 30\(k^3\) = 810

=> \(k^3=810:30\)

=> \(k^3=27\)

=> \(k^3=3^3\)

=> k = 3

=> \(a=2.3=6\)

\(b=3.3=9\)

\(c=5.3=15\)

Vậy .....

b, \(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{c}{9}\)và a - 3b + 4c = 62

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

\(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{c}{9}=\dfrac{a-3b+4c}{4-3.3+4.9}=\dfrac{62}{31}=2\)

=> \(\dfrac{a}{4}=2\Rightarrow a=8\)

\(\dfrac{b}{3}=2\Rightarrow b=6\)

\(\dfrac{c}{9}=2\Rightarrow c=18\)

Vậy .......

c, \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\&a+b+c=49\)

=> \(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}\)

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

\(\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

=> a = 12 . \(\dfrac{3}{2}=18\)

b = 12 . \(\dfrac{4}{3}=16\)

c = 12 . \(\dfrac{5}{4}=15\)

Vậy ...............

d, \(\dfrac{a}{b}=\dfrac{9}{7}\&\dfrac{b}{c}=\dfrac{7}{3},a-b+c=15\)

Ta có : \(\dfrac{a}{b}=\dfrac{9}{7}\Rightarrow\dfrac{a}{9}=\dfrac{b}{7}\)

\(\dfrac{b}{c}=\dfrac{7}{3}\Rightarrow\dfrac{b}{7}=\dfrac{c}{3}\)

=> \(\dfrac{a}{9}=\dfrac{b}{7}=\dfrac{c}{3}\)

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

\(\dfrac{a}{9}=\dfrac{b}{7}=\dfrac{c}{3}=\dfrac{a-b+c}{9-7+3}=\dfrac{15}{5}=3\)

=> \(\dfrac{a}{9}=3\Rightarrow a=27\)

\(\dfrac{b}{7}=3\Rightarrow b=21\)

\(\dfrac{c}{3}=3\Rightarrow c=9\)

Vậy..............

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

Bài này chắc phải giải theo kiểu lớp 7

Giải:

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

\(\dfrac{2a}{3b}=\dfrac{3b}{4c}=\dfrac{4c}{5d}=\dfrac{5d}{2a}=\dfrac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

\(\Rightarrow\left\{{}\begin{matrix}2a=3b\\3b=4c\\4c=5d\\5d=2a\end{matrix}\right.\)\(\Rightarrow2a=3b=4c=5d\)

\(\Rightarrow C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\)

\(=\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}+\dfrac{2a}{2a}\)

\(=1+1+1+1\)

\(=4\)

Vậy \(C=4\)

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 :

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

\(P=\dfrac{5a+10b+15c}{4}+\left(\dfrac{3}{a}+\dfrac{3a}{4}\right)+\left(\dfrac{9}{2b}+\dfrac{b}{2}\right)+\left(\dfrac{4}{c}+\dfrac{c}{4}\right)\)

\(\ge\dfrac{5\left(a+2b+3c\right)}{4}+2\sqrt{\dfrac{3}{a}.\dfrac{3a}{4}}+2\sqrt{\dfrac{9}{2b}.\dfrac{b}{2}}+2\sqrt{\dfrac{4}{c}.\dfrac{c}{4}}\)

\(\Leftrightarrow P\ge\dfrac{5.20}{4}+3+3+2=33\)

Dấu "=" xảy ra khi a=2;b=3;c=4

Vậy \(P_{min}=33\)

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 "

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

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