Tìm các số a, b, c  biết rằng :

     1 . Ta có:       \(\frac{a}{20}=\frac{b}{9}=\frac{c}{6}=\frac{a}{20}=\frac{2b}{9.2}=\frac{4c}{6.4}=\frac{a}{20}=\frac{2b}{18}=\frac{4c}{24}\)

 Ap dụng tính chất dãy tỉ số bắng nhau ta dược :

                    \(\frac{a}{20}=\frac{2b}{18}=\frac{4c}{24}\)=\(\frac{a-2b+4c}{20-18+24}=\frac{13}{26}=\frac{1}{3}\)( do x+2b+4c=13)

Nên : a/20=1/3\(\Leftrightarrow\)     a=1/3.20    \(\Leftrightarrow\)a=20/3

        b/9=1/3   \(\Leftrightarrow\)      b=1/3.9     \(\Leftrightarrow\)    b=3

        c/6=1/3   \(\Leftrightarrow\)      c=1/3.6   \(\Leftrightarrow\)      c= 2

mấy bài sau làm tương tự nhu câu 1

a) 

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

\(\frac{a}{c}=\frac{1}{2}\Rightarrow\frac{a}{1}=\frac{c}{2}\Rightarrow\frac{a}{2}=\frac{c}{4}\)

\(\Rightarrow\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

\(\frac{a^3}{8}=\frac{b^3}{27}=\frac{c^3}{64}=\frac{a^3+b^3+c^3}{8+27+64}=\frac{99}{99}=1\)

Sau đó tính như bình thường thôi bạn

Học tốt~

a, Đặt \(\frac{a}{2}=\frac{b}{3}=\frac{c}{5}=k\)\(\Rightarrow a=2k\); \(b=3k\); \(c=5k\)

Ta có: \(B=\frac{a+7b-2c}{3a+2b-c}=\frac{2k+7.3k-2.5k}{3.2k+2.3k-5k}=\frac{2k+21k-10k}{6k+6k-5k}=\frac{13k}{7k}=\frac{13}{7}\)

b, Ta có: \(\frac{1}{2a-1}=\frac{2}{3b-1}=\frac{3}{4c-1}\)\(\Rightarrow\frac{2a-1}{1}=\frac{3b-1}{2}=\frac{4c-1}{3}\)

\(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{1}=\frac{3\left(b-\frac{1}{3}\right)}{2}=\frac{4\left(c-\frac{1}{4}\right)}{3}\) \(\Rightarrow\frac{2\left(a-\frac{1}{2}\right)}{12}=\frac{3\left(b-\frac{1}{3}\right)}{2.12}=\frac{4\left(c-\frac{1}{4}\right)}{3.12}\)

\(\Rightarrow\frac{\left(a-\frac{1}{2}\right)}{6}=\frac{\left(b-\frac{1}{3}\right)}{8}=\frac{\left(c-\frac{1}{4}\right)}{9}\)\(\Rightarrow\frac{3\left(a-\frac{1}{2}\right)}{18}=\frac{2\left(b-\frac{1}{3}\right)}{16}=\frac{\left(c-\frac{1}{4}\right)}{9}\)

\(\Rightarrow\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}\)

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

\(\frac{3a-\frac{3}{2}}{18}=\frac{2b-\frac{2}{3}}{16}=\frac{c-\frac{1}{4}}{9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-\left(c-\frac{1}{4}\right)}{18+16-9}=\frac{3a-\frac{3}{2}+2b-\frac{2}{3}-c+\frac{1}{4}}{25}\)

\(=\frac{\left(3a+2b-c\right)-\left(\frac{3}{2}+\frac{2}{3}-\frac{1}{4}\right)}{25}=\left(4-\frac{23}{12}\right)\div25=\frac{25}{12}\times\frac{1}{25}=\frac{1}{12}\)

Do đó:  +)  \(\frac{a-\frac{1}{2}}{6}=\frac{1}{12}\)\(\Rightarrow a-\frac{1}{2}=\frac{6}{12}\)\(\Rightarrow a=1\)

+) \(\frac{b-\frac{1}{3}}{8}=\frac{1}{12}\)\(\Rightarrow b-\frac{1}{3}=\frac{8}{12}\)\(\Rightarrow b=1\)

+) \(\frac{c-\frac{1}{4}}{9}=\frac{1}{12}\)\(\Rightarrow c-\frac{1}{4}=\frac{9}{12}\)\(\Rightarrow c=1\)

Gọi k = \(\frac{a-1}{2}=\frac{b+3}{4}=\frac{c-5}{6}\)

=> \(\begin{cases}a=2k+1\\b=4k-3\\c=6k+5\end{cases}\)

=> 5c - 4b - 3a = 30k + 25 - 16k + 12 - 6k - 3 = 8k + 34

=> 8k + 34 = 50

=> k = 2

=> \(\begin{cases}a=5\\b=5\\c=17\end{cases}\)

Ta có :

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{15a-10b}{25}=\frac{6c-15a}{9}\)

\(=\frac{15a-10b+6c-15a}{25+9}=\frac{6c-10b}{34}=\frac{3c-5b}{17}=\frac{5b-3c}{2}\) = 0

=> a+b+c = 5a = - 50 => a = -10; b = -15 ; c = -25