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\(A=\frac{4}{3}.\frac{13}{9}-\frac{4}{3}.\frac{40}{9}\)
\(\Rightarrow A=\frac{4}{3}.\left(\frac{13}{9}-\frac{40}{9}\right)\)
\(\Rightarrow A=\frac{4}{3}.\frac{-27}{9}\)
\(\Rightarrow A=\frac{4}{3}.\left(-3\right)\)
\(\Rightarrow A=-4\)
25% của A là :
\(-4:4=-1\)
Vậy 25% của A là -1
Chúc bạn học tốt !!!
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a) \(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{47.49}=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+......+\frac{2}{47.49}\right)=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{49}\right)=\frac{24}{x+4}\)
\(\Rightarrow\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)
\(\Rightarrow\frac{24}{49}=\frac{24}{x+4}\)
\(\Rightarrow x+4=49\Rightarrow x=45\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\right)=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{47}-\frac{1}{49}\right)=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{49}\right)=\frac{24}{x+4}\Leftrightarrow A=\frac{1}{2}.\frac{48}{49}=\frac{24}{x+4}\)
\(\Rightarrow A=\frac{24}{49}=\frac{24}{x+4}\Leftrightarrow x=49-4=45\)
Bài b) hình như sai đề thì phải đó bạn.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:\(\frac{1}{3x4}+\frac{2}{4x6}+\frac{3}{6x9}+\frac{4}{9x13}+\frac{5}{13x18}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)
\(=\frac{1}{3}-\frac{1}{18}=\frac{6}{18}-\frac{1}{18}=\frac{5}{18}\)
đặt \(A=\frac{1}{3.4}+\frac{2}{4.6}+\frac{3}{6.9}+\frac{4}{9.13}+\frac{5}{13.18}\)=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}=\frac{1}{3}-\frac{1}{18}=\frac{5}{18}\)
=>A=5/18
vậy........
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=\dfrac{3}{1x5}+\dfrac{3}{5x9}+...+\dfrac{3}{121x125}\)
\(\dfrac{4}{3}xE=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{121}-\dfrac{1}{125}\)
\(\dfrac{4}{3}xE=1-\dfrac{1}{125}\)
\(E=\dfrac{124}{125}x\dfrac{3}{4}=\dfrac{93}{125}\)
=52/27-16/3=52/27-144/27=-92/27
\(\dfrac{4}{9}\times\dfrac{13}{3}-\dfrac{4}{3}\times\dfrac{40}{10}\)
\(=\dfrac{52}{27}-\dfrac{16}{3}\)
\(=-\dfrac{92}{27}\)