\(\Rightarrow\frac{5\left(a-1\right)}{5.2}=\frac{3.\left(b+3\right)}{3.4}=\frac{4.\left(c-5\right)}{4.6}\Rightarrow\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}=\frac{5a-5-\left(3b+9\right)-\left(4c-20\right)}{10-12-24}\)

\(\Rightarrow\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}=\frac{5a-5-\left(3b+9\right)-\left(4c-20\right)}{10-12-24}=\frac{\left(5a-3b-4c\right)-5-9+20}{-26}=\frac{46+6}{-26}=-2\)

+) \(\frac{5a-5}{10}=-2\Rightarrow5a-5=-20\Rightarrow a=-3\)

+) \(\frac{3b+9}{12}=-2\Rightarrow3b+9=-24\Rightarrow b=-11\)

+) \(\frac{4c-20}{24}=-2\Rightarrow4c-20=-48\Rightarrow c=-7\)

Vậy ...................

https://qanda.ai/vi/solutions/3C1BcC2xFx

Ta có:

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

\(\Rightarrow\frac{5.\left(a-1\right)}{10}=\frac{3.\left(b+3\right)}{12}=\frac{4.\left(c-5\right)}{24}.\)

\(\Rightarrow\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}\) và \(5a-3b-4c=46.\)

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

\(\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}=\frac{5a-5-3b-9-4c+20}{10-12-24}=\frac{5a-3b-4c+6}{-26}=\frac{\left(5a-3b-4c\right)+6}{-26}=\frac{46+6}{-26}=-2.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a-1}{2}=-2\Rightarrow a-1=-4\Rightarrow a=-3\\\frac{b+3}{4}=-2\Rightarrow b+3=-8\Rightarrow b=-11\\\frac{c-5}{6}=-2\Rightarrow c-5=-12\Rightarrow c=-7\end{matrix}\right.\)

Vậy \(\left(a;b;c\right)=\left(-3;-11;-7\right).\)

Chúc bạn học tốt!

a.S=1+5²+5⁴+...+5²⁰⁰

=>25S=5²+5⁴+5⁶+...+5²⁰²

=>25S-S=(5²+5⁴+5⁶+...+5²⁰²)-(1+5²+5⁴+...+5²⁰⁰)

=>24S=5²⁰²-1

\(\Rightarrow S=\frac{5^{202}-1}{24}\)

b.ta có:

\(\frac{a-1}{2}=\frac{5a-5}{10};\frac{b+3}{4}=\frac{3b+9}{12};\frac{c-5}{6}=\frac{4c-20}{24}\)

\(\Rightarrow\frac{5a-5}{10}=\frac{3b+9}{12}=\frac{4c-20}{24}=\frac{5a-5-3b-9-4c+20}{10-12-24}=\frac{\left(5a-3b-4c\right)+\left(20-9-5\right)}{-26}\)

\(=\frac{46+6}{-26}=\frac{52}{-26}=-2\)

\(\Rightarrow a-1=-2.2=-4\Rightarrow a=-3\)

\(\Rightarrow b+3=-2.4\Rightarrow b=-11\)

\(\Rightarrow c-5=-2.6=-12\Rightarrow c=-7\)

vậy a=-3;b=-11;c=-7

\(\frac{a-1}{2}\) = \(\frac{b+3}{4}\)=\(\frac{c-5}{6}\)và 5a-3b-4c=46

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

a=2k+1 

b= 4k-3

c=6k+5

Thay vào \(\overline{1}\)ta đc : 5(2k+1)-3(4k-3)-4(6k+5)=46

=10k+5-12k-9-32k+20=46

=\(\frac{10k-32k-12k}{5-9-20}=-\frac{46}{24}=-\frac{23}{12}\)??????????????????

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