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Ta có : \(5\frac{8}{17}\div X+\left(-\frac{1}{17}\right)\div X+3\frac{1}{17}\div17\frac{1}{3}=\frac{4}{17}\)
Nên: \(\left(5\frac{8}{17}+\left(-\frac{1}{17}\right)\right)\div X+\frac{52}{17}\div\frac{52}{3}=\frac{4}{17}\)
\(5\frac{7}{17}\div X+\frac{52}{17}\times\frac{3}{52}=\frac{4}{17}\)
\(\frac{92}{17}\div X+\frac{3}{17}=\frac{4}{17}\)
\(\frac{92}{17}\div X=\frac{4}{17}-\frac{3}{17}\)
\(\frac{92}{17}\div X=\frac{1}{17}\)
\(X=\frac{92}{17}\div\frac{1}{17}\)
\(X=92\)
Vậy \(X=92\)
\(5\frac{8}{17}:x+\left(-\frac{1}{17}\right):x=\frac{52}{51}\)
\(\left(5\frac{8}{17}+-\frac{1}{17}\right):x=\frac{52}{51}\)
\(\frac{92}{17}:x=\frac{52}{51}\)
\(X=\frac{92}{17}:\frac{52}{51}=\frac{69}{13}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)
=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)
=> \(2x+\frac{3}{4}=-\frac{7}{4}\)
=> \(2x=\frac{-7}{4}-\frac{3}{4}\)
=> \(2x=-\frac{5}{2}\)
=> \(x=\frac{-5}{2}:2\)
=> \(x=\frac{-5}{4}\)
b) \(\frac{x+1}{3}=\frac{2-x}{2}\)
\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)
\(\Rightarrow2x+2=6-3x\)
\(\Rightarrow2x-3x=6-2\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=4\)
c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)
\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{1}{5}:\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-4x=0\)
Ta có : \(x^2-4x=0\)
\(\Rightarrow xx-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x/5 - 1 = 1/7 : -1/5
2x/5 - 1 = -5/7
2x / 5 = -5/7 + 1
2x / 5 =2/7
=> 7( 2x ) = 2.5
=> 14x = 10
=> x = 10 : 14
=> x = 5/7
\(\frac{-1}{5}\times\left(\frac{2x}{5}-1\right)=\frac{1}{7}\)
\(\frac{2x}{5}-1=\frac{1}{7}\div\left(-\frac{1}{5}\right)\)
\(\frac{2x}{5}-1=\frac{1}{7}\times\left(-5\right)\)
\(\frac{2x}{5}-1=-\frac{5}{7}\)
\(\frac{2x}{5}=-\frac{5}{7}+1\)
\(\frac{2x}{5}=\frac{-5+7}{7}\)
\(\frac{2x}{5}=\frac{2}{7}\)
\(x=\frac{2}{7}\div\frac{2}{5}\)
\(x=\frac{2}{7}\times\frac{5}{2}\)
\(x=\frac{5}{7}\)
\(\frac{3}{x}+\frac{2y}{5}-\frac{1}{5}=0\)
\(\frac{3}{x}+\frac{2y-1}{5}=0\)
\(\frac{3}{x}=\frac{-2y-1}{5}\)
\(x\left(-2y-1\right)=15\)
Tự làm tiếp
Tìm x,y :
\(\frac{3}{x}+\frac{2y}{5}-\frac{1}{5}=0\)
\(\frac{3}{x}+\frac{2y}{5}=0+\frac{1}{5}\)
\(\frac{3}{x}+\frac{2y}{5}=\frac{1}{5}\)
\(\Leftrightarrow x\ne5\)
\(\text{Khi quy đồng để cộng bằng }\frac{1}{5}\text{ ta phỉ quy đồng nên :}\)
\(\frac{3\cdot5}{x\cdot5}+\frac{2y\cdot x}{5\cdot x}=\frac{15}{x\cdot5}+\frac{2y\cdot}{5\cdot x}=\left(\frac{3?}{5\cdot x}>< \frac{4?}{5\cdot x}\right)=\frac{1}{5}\)
\(\text{Ta có 4 trường hợp : }\)
\(\frac{30}{150};\frac{35}{175};\frac{40}{200};\frac{45}{225}\)
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