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1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ : \(x\ne0\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=\frac{-5}{4}\)
\(\left(\frac{-9x}{3x}+\frac{9}{3x}-\frac{x}{3x}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=\frac{-5}{4}\)
\(\frac{-9x+9-x}{3x}:\frac{15+6+10}{15}=\frac{-5}{4}\)
\(\frac{-10x+9}{3x}:\frac{31}{15}=\frac{-5}{4}\)
\(\frac{-10x+9}{3x}=\frac{-31}{12}\)
\(\Leftrightarrow12\left(-10x+9\right)=-31\cdot3x\)
\(\Leftrightarrow-120x+108=-93x\)
\(\Leftrightarrow-120x+93x=-108\)
\(\Leftrightarrow-27x=-108\)
\(\Leftrightarrow x=4\)
b) ĐKXĐ : \(x\ne0\)
\(\frac{-3x}{4}\cdot\left(\frac{1}{x}+\frac{2}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{-3x}{4}=0\\\frac{1}{x}+\frac{2}{7}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\\frac{-2}{-2x}=\frac{-2}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\left(loai\right)\\x=\frac{-7}{2}\end{cases}}\)
Vậy.....
c) phân tích ra rồi làm thôi e :)) a bận rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
làm tiếp cái trước(ấn nhầm)
\(x=\frac{5}{42}-\frac{15}{28}\)
\(x=\frac{5.4}{6.4.7}-\frac{15.6}{4.7.6}\)
\(x=\frac{20}{168}-\frac{90}{168}\)
\(x=\frac{-70}{168}\)
\(x=\frac{-5}{12}\)
2.
1.
\(\frac{11}{13}-\left(\frac{5}{42}-x\right)=-\left(\frac{15}{28}-\frac{11}{13}\right)\)
\(\frac{11}{13}-\frac{5}{42}+x=-\frac{15}{28}+\frac{11}{13}\)
\(\frac{11}{13}-\frac{11}{13}-\frac{5}{42}+\frac{15}{28}=-x\)