a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)

\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)

\(\Leftrightarrow7x-7=12x+18\)

\(\Leftrightarrow5x+18=-7\)

\(\Leftrightarrow5x=-25\)

\(\Leftrightarrow x=-5\)

b) \(\left(2x^2-\frac{1}{2}x\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\left(2x-\frac{1}{2}\right)\left(x^2+1\right)=0\)

Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)

Thay \(a=-\frac{3}{5}\) vào A,ta đc:

\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)

\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)

Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)

\(A=\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6+\frac{7}{4}+\frac{3}{2}\right)\)

\(A=3-\frac{1}{4}+\frac{2}{3}-5+\frac{1}{3}+\frac{6}{5}-6-\frac{7}{4}-\frac{3}{2}\)

\(A=\left(3-5-6\right)-\left(\frac{1}{4}+\frac{7}{4}+\frac{3}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+\frac{6}{5}\)

\(A=-8-\left(2+\frac{3}{2}\right)+1+\frac{6}{5}\)

\(A=-8-2-\frac{3}{2}+1+\frac{6}{5}\)

\(A=-9-\frac{3}{2}+\frac{6}{5}\)

\(A=\frac{-93}{10}\)

Mk lm đc 1 cách thui

Ủng hộ mk nha ^_-

\(4B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n+1\right)}\)

\(4B< \frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{n+1-n}{n\left(n+1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(4B< 1-\frac{1}{n+1}\Rightarrow B< \frac{1}{4}-\frac{1}{4\left(n+1\right)}< \frac{1}{4}\)

Bài làm:

Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)

=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)

Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)

=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)

Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)