Chương 1: MỆNH ĐỀ, TẬP HỢP - Hỏi đáp

\(\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right):\dfrac{456}{123}\)

\(=\dfrac{123}{456}\cdot\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\cdot\dfrac{123}{456}\)

\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}\right)-\left(\dfrac{2009}{2010}-\dfrac{1}{2011}\right)\right]\)

\(=\dfrac{123}{456}\left(\dfrac{2010}{2011}-\dfrac{2011}{2010}-\dfrac{2009}{2010}+\dfrac{1}{2011}\right)\)

\(=\dfrac{123}{456}\left[\left(\dfrac{2010}{2011}+\dfrac{1}{2011}\right)-\left(\dfrac{2011}{2010}+\dfrac{2009}{2010}\right)\right]\)

\(=\dfrac{123}{456}\left(1-2\right)\)

\(=-\dfrac{123}{456}\)

\(\dfrac{5}{8}-\dfrac{3}{8}:\left(-\dfrac{3}{4}\right)-\dfrac{1}{4}\)

\(=\dfrac{5}{8}-\dfrac{3}{8}\cdot\left(-\dfrac{3}{4}\right)-\dfrac{1}{4}\)

\(=\dfrac{5}{8}+2-\dfrac{1}{4}\)

\(=\dfrac{19}{8}\)

m. ng ơi, góc đầu là góc D á là cho nó ra ngoài đầu bên kia nhoa. do mk đánh chữ lộn nó k dính góc c nha

Ta có :

\(xOy+yOx'=180^0\) (2 góc kề bù)

Mà \(xOy=\dfrac{1}{2}yOx'\)

\(\Leftrightarrow yOx'+\dfrac{1}{2}yOx'=180^0\)

\(\Leftrightarrow yOx'\left(1+\dfrac{1}{2}\right)=180^0\)

\(\Leftrightarrow yOx'.1,5=180^0\)

\(\Leftrightarrow yOx'=120^0\)

\(\Leftrightarrow xOy=60^0\)

Ta có :

\(xOy=x'Oy'=60^0\) (đối đỉnh)

\(y'Ox=x'Oy=120^0\) (đối đỉnh)

\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{2450}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{7}-...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(A=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{50}\right)\)

\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+..+\dfrac{1}{50}\right)\)

\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(A=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}=B\)

\(\Rightarrow A:B=1\)

\(18.96=64.27\)

\(\Leftrightarrow\dfrac{18}{64}=\dfrac{96}{27}\)

\(\Leftrightarrow\dfrac{18}{27}=\dfrac{64}{96}\)

\(\Leftrightarrow\dfrac{64}{18}=\dfrac{27}{96}\)

\(\Leftrightarrow\dfrac{27}{18}=\dfrac{96}{64}\)

Ta có : \(18.96=64.27\)

\(\Leftrightarrow\dfrac{96}{64}=\dfrac{27}{18}\)

\(\Leftrightarrow\dfrac{64}{96}=\dfrac{18}{27}\)

\(\Leftrightarrow\dfrac{96}{27}=\dfrac{64}{18}\)

\(\Leftrightarrow\dfrac{27}{96}=\dfrac{18}{64}\)

\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)

\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)

\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)

\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)

\(-\dfrac{1}{30}-\dfrac{1}{15.13}-\dfrac{1}{13.11}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

Đặt \(A=-\dfrac{1}{30}-\dfrac{1}{15.13}-\dfrac{1}{13.11}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)

\(\Rightarrow2A=-\dfrac{1}{15}-\left(\dfrac{2}{15.13}+\dfrac{2}{13.11}+...+\dfrac{2}{5.3}+\dfrac{2}{3.1}\right)\)

\(\Rightarrow2A=-\dfrac{1}{15}-\left(\dfrac{1}{15}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{11}+...+\dfrac{1}{5}-\dfrac{1}{3}+\dfrac{1}{3}-1\right)\)

\(\Rightarrow2A=-\dfrac{1}{15}-\left(\dfrac{1}{15}-1\right)=-\dfrac{1}{15}-\dfrac{1}{15}+1\)

\(=\dfrac{13}{15}\)

Vậy...................

Chúc bạn học tốt!!!

-\(\dfrac{1}{4}\)- -5+\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)-3-\(\dfrac{7}{4}\)+\(\dfrac{4}{3}\)

=-\(\dfrac{1}{4}\)+5+\(\dfrac{1}{3}\)-\(\dfrac{3}{2}\)-3-\(\dfrac{7}{4}\)+\(\dfrac{4}{3}\)

=-(\(\dfrac{1}{4}\)+\(\dfrac{7}{4}\))+(5-3)+(\(\dfrac{1}{3}\)+\(\dfrac{4}{3}\))-\(\dfrac{3}{2}\)

=-2+2+\(\dfrac{5}{3}\)-\(\dfrac{3}{2}\)

=\(\dfrac{1}{6}\)