c, A =a,8+4,b5+0,9c
B = 5a, b6 + 7c
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Áp dụng BĐT AM-GM ta có:
\(A=5a+6b+7c+\frac{1}{a}+\frac{8}{b}+\frac{27}{c}\)
\(=4\left(a+b+c\right)+\left(\frac{1}{a}+a\right)+\left(\frac{8}{b}+2b\right)+\left(\frac{27}{c}+3c\right)\)
\(\ge4\cdot6+2\sqrt{\frac{1}{a}\cdot a}+2\sqrt{\frac{8}{b}\cdot2b}+2\sqrt{\frac{27}{c}\cdot3c}\)
\(\ge24+2+2\cdot4+2\cdot9=52\)
Xảy ra khi \(\frac{1}{a}=a;\frac{8}{b}=2b;\frac{27}{c}=3c\Rightarrow a=1;b=2;c=3\)
a) ta có: \(\frac{a}{4}=\frac{b}{5};\frac{b}{5}=\frac{c}{8}\)
\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{8}=\frac{5a}{20}=\frac{3b}{15}=\frac{3c}{24}\)
ADTCDTSBN
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bn tự áp dụng rùi tìm a;b;c nha
b) ta có: \(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}\)
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có: \(\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}=\frac{3a+9-5b+10+7c-7}{15-15+49}\)
\(=\frac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{49}=\frac{86+12}{49}=\frac{98}{49}=2\)
=>...
c) ta cóL \(\frac{a}{7}=\frac{b}{6}\Rightarrow\frac{a}{35}=\frac{b}{30}\)
\(\frac{b}{5}=\frac{c}{8}\Rightarrow\frac{b}{30}=\frac{c}{48}\)
\(\Rightarrow\frac{a}{35}=\frac{b}{30}=\frac{c}{48}=\frac{2b}{60}\)
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