Ta có :\(A= 2^0+2^1+2^2+2^3+....+ \) \(2^{50}\)

\(2A= 2^1+2^2+2^3+2^4+....+\) \(2^{51}\)

\(2A-A=\left(2^1+2^2+...+2^{51}\right)-\left(2^0+2^1+...+2^{50}\right)\\ A=2^{51}-1\)

Ta có : \(2^{51}-1< 2^{51}\Rightarrow A< B\)

Câu hỏi của nguyen lan anh - Toán lớp 7 | Học trực tuyến

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk;c=dk\)

a) Ta có : \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{ab}{cd}=\dfrac{bk.b}{dk.d}=\dfrac{b^2}{k^2}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)

a) Đặt \(\dfrac{x}{12}=\dfrac{y}{9}=\dfrac{z}{5}=k\)

\(\Rightarrow x=12k;y=9k;z=5k\)

Ta có :\(xyz=12k.9k.5k=20\)

\(540.k^3=20\)

\(\Rightarrow k^3=\dfrac{1}{27}\)

\(\Rightarrow k=\dfrac{1}{3}\)

\(\Rightarrow x=12.\dfrac{1}{3}=4\)

\(\Rightarrow y=9.\dfrac{1}{3}=3\)

\(\Rightarrow z=5.\dfrac{1}{3}=\dfrac{5}{3}\)

Vậy \(x=4 ; y=3 ; z=5/3\)

2. Tính :

\(A=2^0+2^1+2^2+...+2^{100}\\ \Rightarrow2A=2^1+2^2+2^3+...+2^{101}\\ \Rightarrow2A-A=\left(2^0+2^1+2^2+...+2^{100}\right)-\left(2^1+2^2+2^3+...+2^{101}\right)\\ \Rightarrow A=2^{101}-1\)

Ta có : \( 1^2+2^2+3^2+...+10^2=385\)

\(\Rightarrow\) \(2^2( 1^2+2^2+3^2+...+10^2)=385.2^2\)

\(\Leftrightarrow\) \(S=2^2+4^2+6^2...+20^2=385.4\)

\(=> S=1540\)