Bài 1 :
a) \(1-\left(5\frac{3}{8}+x-6\frac{5}{24}\right):12\frac{2}{5}=0\)
\(\Rightarrow\left(\frac{43}{8}+x-\frac{149}{24}\right):\frac{62}{5}=1\)
\(\Rightarrow\left(\frac{129}{24}-\frac{149}{24}\right)+x=\frac{62}{5}\)
\(\Rightarrow\frac{-5}{6}+x=\frac{62}{5}\)
\(\Rightarrow x=\frac{62}{5}-\frac{-5}{6}=\frac{397}{30}\)
Xin lỗi , mình không biết làm phần c bài 1
Bài 2 :
Ta có : \(A=\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{10}\)
\(\Rightarrow7A=-1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^9\)
\(\Rightarrow7A-A=\left[-1+\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^9\right]-\left[\left(-\frac{1}{7}\right)+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{10}\right]\)
\(\Rightarrow6A=-1-\left(\frac{-1}{7}\right)^{10}\Rightarrow A=\frac{-1-\left(\frac{-1}{7}\right)^{10}}{6}\)
Ta có : \(B=\frac{4}{7.31}+\frac{6}{7.41}+\frac{9}{10.41}+\frac{7}{10.57}\)
\(\Rightarrow\frac{1}{5}B=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)
\(\Rightarrow\frac{1}{5}B=\frac{35-31}{35.31}+\frac{41-35}{35.41}+\frac{50-41}{50.41}+\frac{57-50}{50.57}\)
\(\Rightarrow\frac{B}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\Rightarrow\frac{1}{5}B=\frac{1}{31}-\frac{1}{57}=\frac{26}{1767}\)
\(\Rightarrow B=\frac{130}{1767}\)
b) \(1:\left\{-2+1:\left[-2+1:\left(-2+\frac{1}{x}\right)\right]\right\}=\frac{1}{2}\)
\(\Rightarrow-2+1:\left[-2+1:\left(-2+\frac{1}{x}\right)\right]=1:\frac{1}{2}=2\)
\(\Rightarrow1:\left[-2+1:\left(-2+\frac{1}{x}\right)\right]=2-\left(-2\right)=4\)
\(\Rightarrow-2+1:\left(-2+\frac{1}{x}\right)=1:4=\frac{1}{4}\)
\(\Rightarrow1:\left(-2+\frac{1}{x}\right)=\frac{1}{4}-\left(-2\right)=\frac{9}{4}\)
\(\Rightarrow-2+\frac{1}{x}=1:\frac{9}{4}=\frac{4}{9}\)
\(\Rightarrow\frac{1}{x}=\frac{4}{9}-\left(-2\right)=\frac{22}{9}\)
\(\Rightarrow x=\frac{9}{22}\)
