Question

Tìm số hữu tỉỉx,biết:

\(\frac{3x-1}{8}\)+\(\frac{3x+18}{11}\)=\(\frac{3x}{7}\)+\(\frac{3x+20}{13}\)

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Answer 1

Đề bài là tìm số hữu tỉ x nha

Answer 2

\(\frac{3x-1}{8}+\frac{3x+18}{11}=\frac{3x}{7}+\frac{3x+20}{13}\)

\(\Rightarrow\frac{1001\left(3x-1\right)}{8008}+\frac{728\left(3x+18\right)}{8008}=\frac{1144.3x}{8008}+\frac{616\left(3x+20\right)}{8008}\)

\(\Rightarrow3003x-1001+2184x+13104x=3432x+1848x+12320\)\

\(\Rightarrow\)\(19111x=13321\Rightarrow x=\frac{13321}{19111}\)

Answer 3

\(\frac{3\times-1}{8}+\frac{3\times+18}{11}=\frac{3\times}{7}+\frac{3\times+20}{13}\)

\(\Rightarrow\frac{3\times-1}{8}-\frac{3\times}{7}=\frac{3\times+20}{13}-\frac{3\times+18}{11}\)

\(\Rightarrow\frac{3\times-1}{8}+1-\frac{3\times}{7}-1=\frac{3\times+20}{13}-1-\frac{3\times+18}{11}+1\)

\(\Rightarrow\left(\frac{3\times-1}{8}+1\right)-\left(\frac{3\times}{7}+1\right)=\left(\frac{3\times+20}{13}-1\right)-\left(\frac{3\times+18}{11}-1\right)\)

\(\Rightarrow\frac{3\times+7}{8}-\frac{3\times+7}{7}=\frac{3\times+7}{13}-\frac{3\times+7}{11}\)

\(\Rightarrow\left(3\times+7\right)\cdot\left(\frac{1}{8}-\frac{1}{7}\right)=\left(3\times+7\right)\cdot\left(\frac{1}{13}-\frac{1}{11}\right)\)

\(\Rightarrow\left(3\times+7\right)\cdot\left(\frac{-1}{56}\right)=\left(3\times+7\right)\cdot\left(\frac{-2}{143}\right)\)

\(\Leftrightarrow3\times+7=0\)

\(\Leftrightarrow3\times=-7\)

\(\Leftrightarrow\times=\frac{-7}{3}\)

\(\frac{-7}{3}\inℚ\)

Vậy \(\times=\frac{-7}{3}\)