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1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
ta có : x-1/x-5=6/7
=> (x-1).7=(x-5).6
=> 7x-7 = 6x -30
=> 7x-6x = -30+7
=> x= -23
Vậy x=-23 .
\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)
\(\Leftrightarrow7x-7=6x+30\)
hay x=37
a) \(\frac{x+2015}{5}+\frac{x+2015}{6}=\frac{x+2015}{7}+\frac{x+2015}{8}\)
\(\frac{x+2015}{5}+\frac{x+2015}{6}-\frac{x+2015}{7}-\frac{x+2015}{8}=0\)
\(\left(x+2015\right).\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\)
vì \(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\ne0\)
\(\Rightarrow\)x + 2015 = 0
\(\Rightarrow\)x = -2015
b) Tương tự
\(\frac{x-1}{x-5}=\frac{6}{7}\)
\(\Rightarrow\left(x-1\right).7==\left(x-5\right).6\)
\(\Rightarrow7x-7=6x-30\)
\(\Rightarrow7x-6x=-30+7\)
\(\Rightarrow x=-23\)