Kẻ \(Oc//b\)

\(\Rightarrow\widehat{B_1}=\widehat{O_2}\left(SLT\right)\)

\(\Rightarrow\widehat{O_2}=40^o\)

Ta có : a // b

mà Oc // b

\(\Rightarrow\)a // Oc

\(\Rightarrow\widehat{A_1}=\widehat{O_1}\left(SLT\right)\)

\(\Rightarrow\widehat{O_1}=30^o\)

Ta có : \(\widehat{AOB}=\widehat{O_1}+\widehat{O_2}=30^o+40^o=70^o\)

Vậy \(\widehat{AOB}=70^o\)

\(-\frac{2}{5}-\frac{3}{7}=-\frac{29}{35}\)

Đặt \(\frac{a}{2019}=\frac{b}{2020}=\frac{c}{2021}=k\Rightarrow\hept{\begin{cases}a=2019k\\b=2020k\\c=2021k\end{cases}}\)

Khi đó 4(a - b)(b - c) = 4(2019k - 2020k)(2020k - 2021k) = 4(-k)(-k) = 4k² = (2k)² (1)

Lại có (c - a)² = (2021k - 2019k) = (2k)² (2)

Từ (1)(2) => 4(a - b)(b - c) = (c - a)²

Ta có :\(\frac{3z-4y}{2}=\frac{4x-2z}{3}=\frac{2y-3x}{4}\)

=> \(\frac{6z-8y}{4}=\frac{12x-6z}{9}=\frac{8y-12x}{16}=\frac{6z-8y+12x-6z+8y-12x}{4+9+16}=\frac{0}{29}=0\)

=> \(\hept{\begin{cases}3z-4y=0\\4x-2z=0\\2y-3x=0\end{cases}}\Rightarrow\hept{\begin{cases}3z=4y\\4x=2z\\2y=3x\end{cases}}\Rightarrow\hept{\begin{cases}\frac{z}{4}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\\\frac{y}{3}=\frac{x}{2}\end{cases}}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)(đpcm)

tập 1 nhé

TA HAY TOÁN HAY TV

\(\frac{18^3.27^2}{6^2.3^{10}}\)=\(\frac{\left(2.3\right)^2.\left(3^2\right)^3}{\left(2.3\right)^2.\left(3^2\right)^5}\)=\(\frac{2^2.3^2.9^3}{2^2.3^2.9^5}\)=\(\frac{9^3}{9^5}\)=2

Mình giải cho bn r thì bn nhớ k cho mình nhé

Đặt \(S=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)

\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

\(\Rightarrow S< 1\)   (đpcm)

\(\left(\frac{2}{5}+\frac{2}{7}+\frac{2}{11}\right)\div\left(\frac{3}{5}+\frac{3}{7}+\frac{3}{11}\right)\)

\(=2\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)\div3\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)\)

\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{3\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\)

\(=\frac{2}{3}\)

Linh tinh quá!

\(2^{332}=\frac{2^{333}}{2}=\frac{2^{3.111}}{2}=\frac{\left(2^3\right)^{111}}{2}=\frac{8^{111}}{2}\)

\(3^{223}=3^{222}.3=3^{2.111}.3=\left(3^2\right)^{111}.3=9^{111}.3\)

\(9^{111}>8^{111}\Rightarrow3.9^{111}>\frac{8^{111}}{2}\Rightarrow3^{223}>2^{332}\)