Question

a/3 - b=c:3/5 và a b c khác 0 tính Q=2022-(a-5c/5a)^3.(a-3b/b)^3.(5c+3b/3c)^3

Answer 1

\(\frac{a}{3}-b=c:\frac35\)

=>\(\frac{a-3b}{3}=c\cdot\frac53=\frac{5c}{3}\)

=>a-3b=5c

=>a-5c=3b; 5c+3b=a

\(\left(\frac{a-5c}{5a}\right)^3=\left(\frac{3b}{5a}\right)^3=\frac{27b^3}{125a^3}\)

\(\left(\frac{a-3b}{b}\right)^3=\left(\frac{5c}{b}\right)^3=\frac{125c^3}{b^3}\)

\(\left(\frac{5c+3b}{3c}\right)^3=\left(\frac{a}{3c}\right)^3=\frac{a^3}{27c^3}\)

\(Q=2022-\left(\frac{a-5c}{5a}\right)^3\cdot\left(\frac{a-3b}{b}\right)^3\cdot\left(\frac{5c+3b}{3c}\right)^3\)

\(=2022-\frac{27b^3}{125a^3}\cdot\frac{125c^3}{b^3}\cdot\frac{a^3}{27c^3}=2022-1=2021\)