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a ) \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{\left(-11\right)}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{19}{70}\right|\)
=> \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\frac{19}{70}=\frac{3}{35}\)
=> \(\frac{2}{5}+x+\frac{3}{2}=\frac{3}{7}-\frac{3}{35}=\frac{12}{35}\)
=> \(\frac{2}{5}+x=\frac{12}{35}-\frac{3}{2}=-\frac{81}{70}\)
=> \(x=-\frac{81}{70}-\frac{2}{5}=-\frac{109}{70}\)
b) \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
=> \(\frac{3}{4}x-6=\frac{5}{2}\)
=> \(\frac{3}{4}x=\frac{17}{2}\)
=> \(x=\frac{17}{2}:\frac{3}{4}=\frac{34}{3}\)
Câu c,d tự làm nhé
a. \(\frac{3}{7}-\left(\frac{2}{5}+x+\frac{3}{2}\right)=\frac{5}{14}-\left|\frac{4}{35}-\frac{-11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\left(\frac{19}{10}+x\right)=\frac{5}{14}-\left|\frac{4}{35}+\frac{11}{70}\right|\)
\(\Rightarrow\frac{3}{7}-\frac{19}{10}-x=\frac{5}{14}-\left|\frac{19}{70}\right|=\frac{5}{14}-\frac{19}{70}\)
\(\Rightarrow-\frac{103}{70}-x=\frac{3}{35}\)
\(\Rightarrow x=-\frac{103}{70}-\frac{3}{35}\)
\(\Rightarrow x=-\frac{109}{70}\)
b. \(\frac{3}{4}\left(x-8\right)=\frac{5}{7}\left(4-\frac{1}{2}\right)\)
\(\Rightarrow\frac{3}{4}\left(x-8\right)=\frac{5}{7}.\frac{7}{2}=\frac{5}{2}\)
\(\Rightarrow x-8=\frac{10}{3}\)
\(\Rightarrow x=\frac{34}{3}\)
c. \(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\Rightarrow\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
\(\Rightarrow\frac{1}{2}=\frac{2}{3}-7x-4x=\frac{2}{3}-11x\)
\(\Rightarrow11x=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{66}\)
d. \(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)
\(\Rightarrow2-4x-5x+\frac{3}{2}=\frac{7}{4}\)
\(\Rightarrow2-9x=\frac{1}{4}\)
\(\Rightarrow9x=\frac{7}{4}\)
\(\Rightarrow x=\frac{7}{36}\)
1/ \(\frac{1}{3x}:\frac{2}{3}=1\)
<=> \(\frac{3}{3×2×x}=\:1\)
<=> \(\frac{1}{2x}=1\)<=> x = \(\frac{1}{2}\)
a) \(\frac{x}{x+1}=\frac{x+5}{x+7}\)
\(=>x\left(x+7\right)=\left(x+1\right).\left(x+5\right)\)
\(=>x^2+7x=x^2+6x+5\)
\(=>x^2+7x-x^2-6x-5=0\)
\(=>x-5=0\)
\(=>x=5\)
vay \(x=5\)
b) \(\frac{x+7}{x+4}=\frac{x-1}{x-2}\)
\(=>\left(x+7\right)\left(x-2\right)=\left(x+4\right)\left(x-1\right)\)
\(=>x^2+5x-14=x^2+3x-4\)
\(=>x^2+5x-14-x^2-3x+4=0\)
\(=>2x-10=0\)
\(=>2\left(x-5\right)=0\)
\(=>x-5=0\)
\(=>x=5\)
vay \(x=5\)
c) \(\frac{x+2}{x-2}=\frac{x-3}{x+3}\)
\(=>\left(x+2\right)\left(x+3\right)=\left(x-2\right)\left(x-3\right)\)
\(=>x^2+5x+6=x^2-7x+6\)
\(=>x^2+5x+6-x^2+7x-6=0\)
\(=>12x=0\)
\(=>x=0\)
vay \(x=0\)
Theo đề ta có :
\(\frac{x+2}{3}+\frac{x+2}{4}+\frac{x+2}{5}=\frac{x+2}{6}+\frac{x+2}{7}\)
\(\Rightarrow\frac{x+2}{3}+\frac{x+2}{4}+\frac{x+2}{5}-\frac{x+2}{6}-\frac{x+2}{7}=0\)
\(\Rightarrow\left(x+2\right).\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}\right)=0\)
Vì \(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}\ne0\Rightarrow x+2=0\)
\(\Rightarrow x=0-2\Rightarrow x=-2\)
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