\(2a=3b\Rightarrow a=\frac{3b}{2};\frac{b}{10}=\frac{c}{5}\Rightarrow c=\frac{5b}{10}=\frac{b}{2}\) Thay vapf bt a+b-2c=-30 có

\(\frac{3b}{2}+b-\frac{2b}{2}=-30\Rightarrow b=-10\)

Từ đó tìm ra a và c

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2012}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2011}{2012}=\frac{1.2.3....2011}{2.3.4....2012}=\frac{1}{2012}\)

\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2012}\right)\)

\(=\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2011}{2012}\)

\(=\frac{2.3.4...2011}{3.4.5...2012}\)

\(=\frac{1}{1006}\)

b^2=ac ;c^=bd CMR 2b^3+3c^3/2c^3+3d^3=a/d

Ta có :

Q=1/2+1/2^2+1/2^3+...+1/2^100

2Q=1+1/2+1/2^2+...+1/2^99

2Q-Q=(1+1/2+1/2^2+...+1/2^99)-(1/2+1/2^2+1/2^3+...+1/2^100)

Q=1-1/2^100 <1

VẬY Q<1

\(Q=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+.....+\frac{1}{2^{100}}\)

\(\Rightarrow2Q=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}\)

\(\Rightarrow2Q-Q=Q=1-\frac{1}{2^{100}}< 1\)( đpcm )

cả 2 số bằng 2187

a) 3⁵ . 9 = 3⁵ . 3² = 3^{5 + 2} = 3⁷

b) 3⁷ . 2⁰ = 3⁷ . 1 = 3⁷

   CHÚC BẠN HỌC TỐT !! :))