Do \(a;b;c\in Z^+\Rightarrow5^b=a^3+3a^2+5>a+3=5^c\)

\(\Rightarrow5^b>5^c\Rightarrow b>c\)

\(a^3+3a^2+5=5^b\)

\(\Rightarrow a^2\left(a+3\right)+5=5^b\)

\(\Rightarrow a^2\cdot5^c+5=5^b\)

\(\Rightarrow5^b⋮5^c\)

\(\Rightarrow a^2\left(a+3\right)+5⋮a+3\)

\(\Rightarrow5⋮a+3\)

\(\Rightarrow a+3\in\left\{5,1,-1,-5\right\}\)

Mà \(a+b>3\Rightarrow a+3=5\)

\(\Rightarrow a=2\)

\(\Rightarrow b=2;c=1\)

Ta có : \(2a=3b\) \(\Rightarrow\) \(\frac{a}{3}=\frac{b}{2}\) \(\Rightarrow\) \(\frac{a}{21}=\frac{b}{14}\)

           \(5b=7c\) \(\Rightarrow\) \(\frac{b}{7}=\frac{c}{5}\) \(\Rightarrow\) \(\frac{b}{14}=\frac{c}{10}\)

\(\Rightarrow\) \(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a+5c-7b}{63+50-98}=\frac{30}{15}=2\)

                         ( Tính chất dãy tỉ số bằng nhau )

\(\Rightarrow\) \(a=42;b=28;c=20\)

Mục tiêu -1000 sp mong giúp đỡ

Đừng khóa nick nha olm

a: a chia 5 dư 3

=>a=5k+3

b chia 5 dư 2

=>b=5m+2

c chia 5 dư 1

=>c=5x+1

d chia 5 dư 0

=>d=5y

a+b+c+d=5k+3+5m+2+5x+1+5y

=5(k+m+x+y)+6

=5(k+m+x+y+1)+1

=>a+b+c+d chia 5 dư 1

a-b-c-d=5k+3-5m-2-5x-1-5y

=5(k-m-x-y)⋮5

=>a-b-c-d chia 5 dư 0

a-b+c+d=5k+3-5m-2+5x+1+5y

=5(k-m+x+y)+2

=>a-b+c+d chia 5 dư 2

a-b+c-d=5k+3-5m-2+5x+1-5y

=5(k-m+x-y)+2

=>a-b+c-d chia 5 dư 5

b: a+b=5k+3+5m+2=5(k+m+1)

=>a+b chia hết cho 5

c: a+b+d=5k+3+5m+2+5y=5k+5m+5y+5

=5(k+m+y+1)⋮5

2a=3b nên a/3=b/2

5b=7c nên b/7=c/2

=>a/21=b/14=c/4

Áp dụng tính chất của DTSBN, ta đc:

\(\dfrac{a}{21}=\dfrac{b}{14}=\dfrac{c}{4}=\dfrac{5a-7b+5c}{5\cdot21-7\cdot14+5\cdot4}=\dfrac{36}{27}=\dfrac{4}{3}\)

=>a=28; b=56/3; c=16/3

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