ta co b^2=ac

\(\Rightarrow\)a/b=b/c

\(\Rightarrow\)a^2/b^2=b^2/c^2=a/b.b/c

\(\Rightarrow\)a^2+b^2/b^2+c^2=a/c (dpcm)

Vậy ......

Ta có: 3¹+3²+3³+…+3⁹⁹+3¹⁰⁰

=(3¹+3²)+(3³+3⁴)+…+(3⁹⁹+3¹⁰⁰)

=3.(1+3)+3³.(1+3)+…+3⁹⁹.(1+3)

=3.4+3³.4+…+3⁹⁹.4

=(3+3³+…+3⁹⁹).4 chia hết cho 4

=>3¹+3²+3³+…+3⁹⁹+3¹⁰⁰ chia hết cho 4

Đặt   \(A=3+3^2+3^3+...+3^{99}+3^{100}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)

\(=3\left(1+3\right)+3^{ 3}\left(1+3\right)+...+3^{99}\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+...+3^{99}\right)\)

\(=4\left(3+3^3+...+3^{99}\right)\)

Vì 4 chia hết cho 4 nên \(4\left(3+3^3+...+3^{99}\right)\)

Vậy A chia hết cho 4

từ giả thiết:

b^2=ac;c^2=bd =>\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

ta có: \(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)

lại có:

\(\frac{a^3}{b^3}=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a.b.c}{b.c.d}=\frac{a}{d}\left(2\right)\)

từ 1 và 2=>đpcm

b;c;d thoả mãn b 2 =ac; c - Giúp tôi giải toán - Hỏi đáp, thảo ... nho lik e vao do dug 10000000000000000000%

\(\frac{3^2.3^8}{27^3}=3x=>\frac{3^{10}}{\left(3^3\right)^3}=3x=>\frac{3^{10}}{3^9}=3x=>3^{10-9}=3x=>3x=3=>x=1\)