Ta có: \(\left(-2a^2b^3\right)^{10}=\left(-2\right)^{10}.\left(a^2\right)^{10}.\left(b^3\right)^{10}=2^{10}.a^{20}.b^{30}\)

\(\left(3b^2c^4\right)^{15}=3^{15}.\left(b^2\right)^{15}.\left(c^4\right)^{15}=3^{15}.b^{30}.c^{60}\)

Vì \(2^{10}.a^{20}.b^{30}\ge0\) với mọi a;b

  \(3^{15}.b^{30}.c^{60}\ge0\) với mọi b;c

=>\(2^{10}.a^{20}.b^{30}+3^{15}.b^{30}.c^{60}\ge0\) với mọi a;b;c

Mà \(2^{10}.a^{20}.b^{30}+3^{15}.b^{30}.c^{60}=0\) (theo đề)

=>\(2^{10}.a^{20}.b^{30}=3^{15}.b^{30}.c^{60}=0\)

=>a²⁰.b³⁰=b³⁰.c⁶⁰=0

=>a.b=b.c=0

Vậy b=0;a và c tùy ý hoặc a=c=0;b tùy ý

Ta có: \(C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\)

\(=\dfrac{2a}{3b}\cdot4=\dfrac{8a}{3b}\)

c) Vì  \(\dfrac{2a}{3b}=\dfrac{3b}{4c}=\dfrac{4c}{5d}=\dfrac{5d}{2a}\) nên theo t/c của DTSBN ta có :

\(\Rightarrow\)\(C=\dfrac{2a}{3b}+\dfrac{3b}{4c}+\dfrac{4c}{5d}+\dfrac{5d}{2a}\) = \(\dfrac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

ko trả lời đâu bạn ơi

Theo t/c dãy tỉ số=nhau:

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

Khi đó \(C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=1+1+1+1=4\)

Vậy C=4

C = 4 đó bạn

very good

Ta có : \(\frac{a}{3}\)=\(\frac{b}{4}\)=>\(\frac{2a}{18}\)=\(\frac{3b}{36}\)

             \(\frac{b}{3}=\frac{c}{5}=>\frac{3b}{36}=\frac{c}{20}\)

=>\(\frac{2a}{18}=\frac{3b}{36}=\frac{c}{20}\)=\(\frac{2a-3b+c}{18-36+20}\)=\(\frac{6}{2}\)=3

\(\frac{2a}{18}\)=3  =>  a=24

\(\frac{3b}{36}\)=3  =>  b=36

\(\frac{c}{20}\)=3  =>  c=60

   Vậy a=24 ; b=36 ; c=60