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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: \(\frac{a}{b}=\frac{c}{d}\)\(\Leftrightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)(1)
\(\Leftrightarrow\frac{a^2}{b^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{c}{d}=\frac{ac}{bd}\)(2)
Từ (1) và (2) => \(\frac{a^2+c^2}{b^2+d^2}=\frac{ac}{bd}\)
b, Ta có: \(\frac{a}{b}=\frac{c}{d}\)\(\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)\(\Leftrightarrow\frac{4a}{4c}=\frac{3b}{3d}=\frac{4a+3b}{4c+3d}=\frac{4a-3b}{4c-3d}\)
\(\Leftrightarrow\frac{4a+3b}{4c+3d}=\frac{4a-3b}{4c-3d}\)
\(\Leftrightarrow\left(4a+3b\right)\left(4c-3d\right)=\left(4a-3b\right)\left(4c+3d\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\frac{a}{2}\)\(=\)\(\frac{b}{5}\)\(=\)\(\frac{c}{7}\)\(=\)K
=> a=2K
b=5k
c=7k
=> \(\frac{4a+2b-c}{a-b-c}\)= \(\frac{8k+10k-7k}{2k-5k-7k}\)= \(\frac{k.\left(8+10-7\right)}{k.\left(2-5-7\right)}\)= \(\frac{8+10-7}{2-5-7}\)= \(\frac{-11}{10}\)
Đặt \(\frac{a}{2}=\frac{b}{5}=\frac{c}{7}=k\)
\(\Rightarrow a=2k\)
\(b=5k\)
\(c=7k\)
\(\Rightarrow\frac{4a+2b-c}{a-b-c}\)
\(=\frac{4\left(2k\right)+2\left(5k\right)-7k}{2k-5k-7k}\)
\(=\frac{\left(8+10-7\right)k}{\left(2-5-7\right)k}\)
\(=-\frac{11}{10}\)
Vậy ...